- When two factors both contain \(i\), what power of \(i\) appears? - Multiply the ordinary number factors separately. - Keep the negative sign in part 3 when multiplying.

1. \(i\cdot i=i^2=-1\). 2. \((3i)(2i)=6i^2=-6\). 3. \((-i)(4i)=-4i^2=4\).

1) \(-1\) 2) \(-6\) 3) \(4\)

- Divide each term in a sum by the real divisor. - Cancel a common nonzero factor of \(i\) in the numerator and denominator. - Treat \(i\) as a factor when no higher power of \(i\) is created.

1. Divide both parts by \(6\): \((18+6i)\div6=3+i\). 2. \(20i\div5i=4\). 3. Divide both parts by \(4\): \((-12+4i)\div4=-3+i\). 4. \(7i\div i=7\).

1) \(3+i\) 2) \(4\) 3) \(-3+i\) 4) \(7\)

- Standard rectangular form places the real term before the term containing \(i\). - The imaginary part is a real number: it is the coefficient of \(i\). - Keep the sign attached to each component.

1. Reorder the terms to write \(z=7-3i\). 2. The real part is the coefficient-free real term, \(7\). 3. The imaginary part is the coefficient of \(i\), which is \(-3\).

\(z=7-3i\). Its real part is \(7\), and its imaginary part is \(-3\).

- Read the horizontal coordinate first. - Read the vertical coordinate as the coefficient of \(i\). - Write the result in \(a+bi\) form.

1. The horizontal coordinate gives the real part and the vertical coordinate gives the imaginary part. 2. Point \(A\) is at \((-3,2)\), so the represented complex number is \(-3+2i\).

\(-3+2i\)

- What changes, and what stays the same, when a complex number is conjugated? - Geometrically, conjugation reflects a point across one coordinate axis. - Compare the coordinates of \(z\) with each candidate point.

1. The conjugate keeps the real part and changes the sign of the imaginary part. 2. Point \(z\) is at \((2,3)\), so its conjugate is represented by \((2,-3)\). 3. Candidate \(A\) is at \((2,-3)\), so \(A\) represents \(\overline{z}\).

Point \(A\)

- What happens when you multiply two negative real numbers? - Consider the sign of the square of any real number.

1. The square of any real number is nonnegative, so no real square equals \(-1\). 2. The imaginary unit \(i\) is defined by \(i^2=-1\).

Every real square is nonnegative, so no real number can satisfy \(x^2=-1\).

- Separate the positive square factor from the negative sign. - What symbol represents a square root of \(-1\)?

1. Since \(-25=25\cdot(-1)\), \(\sqrt{-25}=\sqrt{25}\sqrt{-1}\). 2. By definition, \(\sqrt{-1}=i\), so \(\sqrt{-25}=5i\).

\(5i\)

- What does raising a nonzero number to the zero power mean? - Recall what the first power does. - What single equation defines \(i\)?

1. Any nonzero number to the power \(0\) is \(1\), so \(i^0=1\). 2. Raising to the first power leaves the number unchanged, so \(i^1=i\). 3. By definition, \(i^2=-1\).

a) \(1\) b) \(i\) c) \(-1\)

- Treat the real terms and the terms containing \(i\) separately. - What coefficient does the second \(i\) have?

1. Add real parts: \(2+4=6\). 2. Add imaginary parts: \(3i+i=4i\). 3. The sum is \(6+4i\).

\(6+4i\)

- Identify the real term and the term containing \(i\). - If you take the conjugate of your result, do you get back the starting number?

1. A complex conjugate keeps the real part and changes the sign of the imaginary part. 2. Thus \(\overline{z}=5+3i\). Only the imaginary part changes sign.

\(\overline{z}=5+3i\); the imaginary part changes sign.

- Combine the real parts separately from the terms containing \(i\). - After adding, check which part of the result is zero. - A purely imaginary number has zero real part and a nonzero imaginary part.

1. Add the real parts: \(4+(-4)=0\). 2. Add the imaginary parts: \(-2i+5i=3i\). 3. The result is \(0+3i=3i\), which is purely imaginary.

\(0+3i=3i\); purely imaginary.

- Group the real terms together. - Treat the terms containing \(i\) as like terms. - Make sure both parts of the second complex number are subtracted.

1. Subtract the real parts: \(4-2=2\). 2. Subtract the imaginary parts: \(3i-i=2i\). 3. The difference is \(2+2i\).

\(2+2i\)

- Which real number produces a \(180^\circ\) rotation when used as a complex multiplier? - How is a dilation represented by multiplication? - Combine the two transformations into one multiplier.

1. A \(180^\circ\) rotation about the origin is multiplication by \(-1\). 2. A dilation by a factor of \(3\) is multiplication by \(3\). 3. The combined transformation is multiplication by \(w=3\cdot(-1)=-3\). 4. Therefore, \(z_2=-3(1+2i)=-3-6i\).

\(z_2=-3-6i\) and \(w=-3\)

- Add or subtract corresponding real and imaginary parts. - A complex number \(a+bi\) corresponds to the point \((a, b)\). - Use the magnitude formula for the difference. - Think about what the length of the segment between two points represents.

1. \(s=(4-2)+(2+3)i=2+5i\). 2. \(d=(4-(-2))+(2-3)i=6-i\). 3. The endpoint for \(s\) is \((2, 5)\), and the endpoint for \(d\) is \((6, -1)\). 4. \(|z_1-z_2|=|6-i|=\sqrt{6^2+(-1)^2}=\sqrt{37}\approx6.08\). 5. The magnitude of the difference is the distance between the points represented by \(z_1\) and \(z_2\).

a) \(s=2+5i\); \(d=6-i\) b) \((2, 5)\) and \((6, -1)\) c) \(|z_1-z_2|=\sqrt{37}\approx6.08\); this is the distance between the two points.

- Use \(i^2=-1\). - Multiply the binomials carefully. - To remove a complex number from the denominator, multiply by its conjugate. - Use the difference-of-squares pattern in the denominator.

1. \((4-3i)(1+2i)=4+8i-3i-6i^2=10+5i\). 2. For the quotient, multiply by the conjugate of the denominator: \(\frac{4-3i}{1+2i}\cdot\frac{1-2i}{1-2i}\). 3. The denominator is \((1+2i)(1-2i)=1+4=5\). 4. The numerator is \((4-3i)(1-2i)=4-8i-3i+6i^2=-2-11i\). 5. Therefore, \(\frac{z_1}{z_2}=-\frac{2}{5}-\frac{11}{5}i\).

a) \(10+5i\) b) \(-\frac{2}{5}-\frac{11}{5}i\)

- Combine real parts with real parts and imaginary parts with imaginary parts. - Distribute a minus sign carefully across parentheses. - Use \(i^2=-1\) to simplify higher powers of \(i\). - Look for a binomial-square pattern.

1. Combine real and imaginary parts: \((12-7-5)+(-5+8-3)i=0\). 2. Since \(i^3=-i\), \(4(0.5-i)-(3+2i)+i^3=2-4i-3-2i-i=-1-7i\). 3. \((2+i)^2=4+4i+i^2=3+4i\), so \((3+4i)-(4+4i)=-1\).

1) \(0\) 2) \(-1-7i\) 3) \(-1\)

- Write both sides in the form \(a+bi\). - Two complex numbers are equal when their real parts match and their imaginary parts match. - Simplify \(i^2\) first. - Set corresponding parts equal.

1. Write the left side as \((x+2)+(y-3)i\). 2. Since \(i^2=-1\), the right side is \(-5+4i\). 3. Equate real parts: \(x+2=-5\), so \(x=-7\). 4. Equate imaginary parts: \(y-3=4\), so \(y=7\).

\(x=-7\) and \(y=7\)

- Add corresponding real and imaginary parts. - Interpret each complex number as a vector from the origin. - Think about the parallelogram rule for vector addition. - Recall the property that allows addends to switch order.

1. \(s_A=(4-1)+(2+3)i=3+5i\), and \(s_B=(-1+4)+(3+2)i=3+5i\). 2. Complex-number addition corresponds to vector addition. The origin, the two original points, and the sum point form a parallelogram. Because both orders give the same sum point, the diagram illustrates the commutative property: \(z_1+z_2=z_2+z_1\).

1. \(s_A=3+5i\) and \(s_B=3+5i\) 2. The points form a parallelogram and illustrate the commutative property of addition.

- Subtract corresponding vector components. - Use the magnitude formula for \(a+bi\). - Interpret the magnitude of a difference vector.

1. \(w=(7-3)+(5-2)i=4+3i\). 2. \(|w|=\sqrt{4^2+3^2}=5\). 3. Because \(w=z_1-z_2\), \(|w|=|z_1-z_2|\) is the distance between the points represented by \(z_1\) and \(z_2\). The distance is \(5\) units.

1. \(w=4+3i\) 2. \(|w|=5\) 3. The two points are \(5\) units apart.

- Identify the real and imaginary parts separately. - When subtracting parentheses, reverse the signs inside. - Check whether any terms cancel. - Combine the \(u\)-terms and the \(vi\)-terms separately.

1. For part a, combine the real parts: \(3.5-1.5-0.2=1.8\). Combine the imaginary coefficients: \(-0.4+0.4-(-1.75)=1.75\). Thus, the result is \(1.8+1.75i\). 2. For part b, distribute and combine like terms: \(4u-5vi-2u-vi-u+3vi=u-3vi\).

a) \(1.8+1.75i\) b) \(u-3vi\)

- Pay close attention to signs when removing parentheses preceded by a minus sign. - Combine terms with \(i\) separately from terms without \(i\). - Work from the innermost grouping outward.

1. Simplify inside the brackets: \((2a+bi)-(a-5bi)+(4a+2bi)=2a+bi-a+5bi+4a+2bi=5a+8bi\). 2. Subtract this result: \((3a-4bi)-(5a+8bi)=-2a-12bi\).

\(-2a-12bi\)

- Combine terms containing \(i\) separately from terms without \(i\). - Pay attention to the minus sign before parentheses. - Use common denominators when combining fractions.

1. Combine the real parts: \(4.5-1.8+0.3=3\). Combine the imaginary coefficients: \(-1.2-2.4-0.9=-4.5\). The result is \(3-4.5i\). 2. The real part is \(\frac{3}{4}+\frac{1}{2}-1=\frac{1}{4}\). The imaginary coefficient is \(\frac{5}{6}-\frac{1}{3}+\frac{1}{4}=\frac{3}{4}\). The result is \(\frac{1}{4}+\frac{3}{4}i\).

1) \(3-4.5i\) 2) \(\frac{1}{4}+\frac{3}{4}i\)

- Replace \(i^2\) with \(-1\). - Multiply complex binomials as you would real binomials. - For the fourth power, square in stages. - Track the sign of every \(i^2\) term.

1. \((4-3i)(2+5i)=8+20i-6i-15i^2=23+14i\). 2. \((1+i)^2=2i\), so \((1+i)^4=(2i)^2=-4\). 3. \((3-0.5i)(2+4i)=6+12i-i-2i^2=8+11i\). 4. \((\sqrt{7}+3i)(\sqrt{7}-3i)=7-(3i)^2=16\).

1) \(23+14i\) 2) \(-4\) 3) \(8+11i\) 4) \(16\)

- Form the conjugate by changing the sign of the imaginary part. - Use the difference-of-squares pattern. - Replace \(i^2\) with \(-1\). - Recall the possible values of squares of real numbers.

1. \(\overline{z_1}=6+8i\), so \(z_1\overline{z_1}=(6-8i)(6+8i)=6^2+8^2=100\). 2. \(\overline{z_2}=\sqrt{7}-i\sqrt{2}\), so \(z_2\overline{z_2}=7+2=9\). 3. In general, \((a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2\). 4. Since squares of real numbers are nonnegative, \(a^2+b^2\) is real and at least \(0\).

a) \(z_1\overline{z_1}=100\) and \(z_2\overline{z_2}=9\) b) \(z\overline{z}=a^2+b^2\ge0\)

- Make the denominator real by multiplying by its conjugate. - Use the product of conjugates in the denominator. - Replace \(i^2\) with \(-1\). - Simplify the real and imaginary coefficients separately.

1. For part a, multiply by the conjugate of the denominator: \(\frac{3-4i}{2+i}\cdot\frac{2-i}{2-i}\). 2. The denominator is \(5\), and the numerator is \((3-4i)(2-i)=2-11i\). Thus, the result is \(\frac{2}{5}-\frac{11}{5}i\). 3. For part b, \(\frac{5}{1-2i}\cdot\frac{1+2i}{1+2i}=\frac{5+10i}{5}=1+2i\).

a) \(\frac{2}{5}-\frac{11}{5}i\) b) \(1+2i\)

- Multiply by the conjugate of the denominator to make it real. - Use the difference-of-squares pattern for conjugate products. - Replace \(i^2\) with \(-1\) when expanding the numerator.

1. Multiply by \(\frac{1-i}{1-i}\): \(\frac{(2+4i)(1-i)}{2}=\frac{6+2i}{2}=3+i\). 2. Multiply by \(\frac{1+3i}{1+3i}\): \(\frac{10(1+3i)}{10}=1+3i\). 3. Multiply by \(\frac{2-i}{2-i}\): \(\frac{5i(2-i)}{5}=\frac{5+10i}{5}=1+2i\).

1) \(3+i\) 2) \(1+3i\) 3) \(1+2i\)

- Multiply by the conjugate of the denominator. - Use \(i^2=-1\). - Look for a common factor in the numerator and denominator. - A pure multiple of \(i\) has real part zero.

1. Multiply by the conjugate of the denominator: \(z=\frac{(a+2i)(2+ai)}{(2-ai)(2+ai)}\). 2. The denominator is \(4+a^2\). 3. The numerator is \(2a+a^2i+4i+2ai^2=i(a^2+4)\). 4. Since \(a^2+4\ne0\), \(z=i\). 5. Therefore, \(\operatorname{Re}(z)=0\) and \(\operatorname{Im}(z)=1\).

\(\operatorname{Re}(z)=0\) and \(\operatorname{Im}(z)=1\)

- Multiply by a conjugate to remove \(i\) from the denominator. - Compare a complex number with its conjugate. - Recall that the reciprocal of \(z\) is \(\frac{1}{z}\).

1. For part a, multiply by the conjugate of the denominator: \(\frac{3+i}{1-2i}\cdot\frac{1+2i}{1+2i}\). 2. The numerator is \(1+7i\), and the denominator is \(5\), so \(w=\frac{1}{5}+\frac{7}{5}i\). 3. For part b, \(z_1^{-1}=\frac{1}{3+i}\cdot\frac{3-i}{3-i}=\frac{3-i}{10}=\frac{3}{10}-\frac{1}{10}i\).

a) \(w=\frac{1}{5}+\frac{7}{5}i\) b) \(z_1^{-1}=\frac{3}{10}-\frac{1}{10}i\)

- Rewrite each quotient with a real denominator. - Use \(i^2=-1\). - A conjugate product is real. - Simplify the entire expression to \(a+bi\).

1. Rationalize the first denominator: \(\frac{1+3i}{1-i}\cdot\frac{1+i}{1+i}=\frac{-2+4i}{2}=-1+2i\). 2. \(\frac{2}{i}=\frac{2i}{i^2}=-2i\). 3. Therefore, \(z=(-1+2i)-(-2i)=-1+4i\). 4. Thus, \(\operatorname{Re}(z)=-1\) and \(\operatorname{Im}(z)=4\).

\(\operatorname{Re}(z)=-1\) and \(\operatorname{Im}(z)=4\)

- Apply the binomial-square formulas. - Replace \(i^2\) with \(-1\). - Distribute the subtraction sign carefully. - You may also use the difference-of-squares identity.

1. \((2+3i)^2=4+12i+9i^2=-5+12i\). 2. \((2-3i)^2=4-12i+9i^2=-5-12i\). 3. Subtract: \((-5+12i)-(-5-12i)=24i\).

\(24i\)

- List the first four powers of \(i\) and look for a pattern. - Find the remainder when each exponent is divided by \(4\). - Combine the real parts and imaginary parts after simplifying.

1. The powers of \(i\) repeat every four: \(i,-1,-i,1\). 2. Dividing \(2023\) by \(4\) leaves remainder \(3\), so \(i^{2023}=-i\). 3. Dividing \(2024\) by \(4\) leaves remainder \(0\), so \(i^{2024}=1\). 4. Dividing \(2025\) by \(4\) leaves remainder \(1\), so \(i^{2025}=i\). 5. Thus \(z=-i+1+i=1+0i\).

\(z=1+0i\)

- Look at the repeating four-power cycle for \(i\). - The remainder after division by \(4\) identifies each power in the cycle. - For a power of \(-i\), also consider the negative sign.

1. The powers of \(i\) repeat every four. Since \(123\) leaves remainder \(3\) when divided by \(4\), \(i^{123}=-i\). 2. The four terms in part b are \(-1,-i,1,i\), which sum to \(0\). 3. Since \((-i)^7=i\) and \(i^{21}=i\), their product is \(i^2=-1\).

a) \(-i\) b) \(0\) c) \(-1\)

- Multiply by the conjugate of the denominator. - Use the difference-of-squares pattern. - Replace \(i^2\) with \(-1\). - Divide both parts of the numerator by the real denominator.

1. Multiply by the conjugate of the denominator: \(z=\frac{(3-i\sqrt{2})(1-i\sqrt{2})}{(1+i\sqrt{2})(1-i\sqrt{2})}\). 2. The denominator is \(1+2=3\). 3. The numerator is \(3-3i\sqrt{2}-i\sqrt{2}+2i^2=1-4i\sqrt{2}\). 4. Therefore, \(z=\frac{1}{3}-\frac{4\sqrt{2}}{3}i\).

\(z=\frac{1}{3}-\frac{4\sqrt{2}}{3}i\)

- Modulus can be interpreted as distance from the origin. - Use each point's horizontal and vertical coordinates as the legs of a right triangle. - Compare squared distances first if that is simpler.

1. Modulus is the distance from the origin. 2. For \(p=(3,4)\), \(|p|=\sqrt{3^2+4^2}=5\). 3. For \(q=(-4,1)\), \(|q|=\sqrt{(-4)^2+1^2}=\sqrt{17}\). 4. For \(r=(0,-5)\), \(|r|=5\). 5. Therefore, \(|q|<|p|=|r|\).

\(|q|<|p|=|r|\), with \(|q|=\sqrt{17}\) and \(|p|=|r|=5\).

- What happens when you subtract an entire expression in parentheses? - After removing parentheses, combine real terms and imaginary terms separately.

1. Distribute the minus sign: \(7-2i-3-5i\). 2. Combine real terms and imaginary terms: \((7-3)+(-2-5)i=4-7i\).

\(4-7i\)

- Pay attention to the order of subtraction. - Distribute each subtraction sign through the entire second number. - Compare the real and imaginary parts in the two answers.

1. \(u-v=(2-i)-(-3+4i)=5-5i\). 2. \(v-u=(-3+4i)-(2-i)=-5+5i\). 3. These are opposites: \(v-u=-(u-v)\).

a) \(u-v=5-5i\); \(v-u=-5+5i\) b) They are opposites.

- To get the next power, multiply the previous one by \(i\). - Replace \(i^2\) with \(-1\) when it appears. - Compare the last answer with the earlier powers.

1. Multiply \(i^2=-1\) by \(i\) to get \(i^3=-i\). 2. Multiplying again gives \(i^4=(-i)i=-i^2=1\). 3. Finally, \(i^5=i\), so the value \(i\) repeats after four powers.

\(i^3=-i\), \(i^4=1\), \(i^5=i\). The value \(i\) repeats.

- Multiply each term in the first pair of parentheses by each term in the second. - Replace any square of \(i\) with its value. - Combine real terms and imaginary terms separately.

1. Distribute: \((1+2i)(3+i)=3+i+6i+2i^2\). 2. Since \(i^2=-1\), this becomes \(3+7i-2=1+7i\).

\(1+7i\)

- A squared binomial is a product of two equal binomials. - Write the full distribution before combining terms. - Simplify \(i^2\) only after including the cross-products.

1. The student squared each term but left out the two cross-products. 2. Expanding gives \((2+i)^2=4+2i+2i+i^2\). 3. Since \(i^2=-1\), the correct result is \(3+4i\). The missing term is \(4i\).

The missing term is \(4i\); \((2+i)^2=3+4i\).

- Expand the two factors before simplifying powers of \(i\). - Compare the two mixed terms in the expansion. - What happens to the imaginary terms when their signs differ?

1. Multiply: \((2+i)(2-i)=4-2i+2i-i^2\). 2. The imaginary terms cancel. Using \(i^2=-1\) gives \(4-(-1)=5\). 3. Conjugates have opposite imaginary parts, so their product has no imaginary term.

\(5\). The imaginary terms cancel when conjugates are multiplied.

- A quotient in \(a+bi\) form should have a real denominator. - Which factor would remove the imaginary part of this denominator? - Simplify the real denominator before separating the two parts.

1. Multiply numerator and denominator by the conjugate \(1-i\). 2. This gives \(\frac{1-i}{(1+i)(1-i)}=\frac{1-i}{1-i^2}=\frac{1-i}{2}\). 3. In \(a+bi\) form, the result is \(\frac{1}{2}-\frac{1}{2}i\).

\(\frac{1}{2}-\frac{1}{2}i\)

- The horizontal coordinate is the real part and the vertical coordinate is the imaginary part. - Add the real and imaginary parts separately. - Convert the result back to a pair of coordinates.

1. The point has coordinates \((2,-3)\), so \(z=2-3i\). 2. Adding \(-1+4i\) gives \((2-1)+(-3+4)i=1+i\). 3. The resulting point is \((1,1)\).

\(1+i\); point \((1,1)\)

- Combine the real terms and the terms containing \(i\) separately. - What must the real part equal for the sum to be purely imaginary? - Check both parts of the result using your value of \(k\).

1. Add the real parts and the imaginary parts: \((k+3i)+(4-2i)=(k+4)+i\). 2. A purely imaginary number has real part \(0\), so \(k+4=0\) and \(k=-4\). 3. The sum is \(i\).

\(k=-4\); the sum is \(i\).

- Think about the subtraction sign in front of the entire second number. - Rewrite the expression without parentheses before combining like terms. - Compare the sign of each imaginary term with Mateo's result.

1. Subtract the whole second complex number: \((4-3i)-(1-2i)=4-3i-1+2i\). 2. Combining like terms gives \(3-i\). 3. Mateo treated the imaginary term \(-2i\) as if it were being added rather than subtracted.

Mateo did not change the sign of the \(-2i\) term when subtracting. The correct difference is \(3-i\).

- In a conjugate pair, which coefficient changes sign? - Use the given negative imaginary part to determine the unknown coefficient. - Add the two numbers and check which terms cancel.

1. Conjugates have equal real parts and opposite imaginary parts, so \(k=6\). 2. Their sum is \((4+6i)+(4-6i)=8\). 3. The imaginary terms cancel, leaving a real number.

a) \(k=6\) b) The sum is \(8\); the opposite imaginary parts cancel.

- What factor could make the denominator a difference of squares? - Keep the same multiplier in the numerator and denominator. - Expand each product and combine real and imaginary terms separately.

1. Multiply the numerator and denominator by \(2+i\), the conjugate of \(2-i\). 2. The denominator is \((2-i)(2+i)=4-i^2=5\). 3. The numerator is \((3+i)(2+i)=6+3i+2i+i^2=5+5i\). 4. Dividing by \(5\) gives \(1+i\).

\(1+i\). The denominator is real because the imaginary cross terms cancel in \((2-i)(2+i)=5\).

- Multiply complex numbers as you would multiply binomials, then replace \(i^2\) with \(-1\). - To make a complex denominator real, multiply by its complex conjugate. - The conjugate changes only the sign of the imaginary part. - Use the difference-of-squares pattern.

1. Multiply and use \(i^2=-1\): \((1+2i)(3-i)=3-i+6i-2i^2=5+5i\). 2. Multiply the numerator and denominator by the conjugate of the denominator: \(\frac{1+2i}{3-i}\cdot\frac{3+i}{3+i}=\frac{(1+2i)(3+i)}{(3-i)(3+i)}\). 3. The numerator is \(3+i+6i+2i^2=1+7i\), and the denominator is \(9-i^2=10\). 4. Thus, \(\frac{z_1}{z_2}=\frac{1}{10}+\frac{7}{10}i\). 5. In general, \((a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2\). Because \(a^2\ge0\) and \(b^2\ge0\), this is a nonnegative real number.

a) \(z_1z_2=5+5i\) b) \(\frac{z_1}{z_2}=\frac{1}{10}+\frac{7}{10}i\) c) \(z\overline{z}=a^2+b^2\ge0\)

- Opposite sides of a parallelogram represent equal vectors. - Express the displacement from one point to another as a difference. - Add the vector from \(B\) to \(C\) to point \(A\).

1. In parallelogram \(ABCD\), \(\overrightarrow{AD}=\overrightarrow{BC}\). 2. The vector from \(B\) to \(C\) is \(z_C-z_B=(2+5i)-(4+2i)=-2+3i\). 3. Add this vector to \(z_A\): \(z_D=z_A+(z_C-z_B)=(1+i)+(-2+3i)=-1+4i\).

\(z_D=-1+4i\)

- Think about vector addition using the parallelogram rule. - The magnitude of a complex number is its distance from the origin. - Compare the three vector lengths as sides of a triangle.

1. Add corresponding parts: \(s=(3-1)+(2+4)i=2+6i\). Geometrically, the vector for \(s\) is the diagonal of the parallelogram formed by the vectors for \(z_1\) and \(z_2\). 2. The magnitudes are \(|z_1|=\sqrt{3^2+2^2}=\sqrt{13}\), \(|z_2|=\sqrt{(-1)^2+4^2}=\sqrt{17}\), and \(|s|=\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}\). 3. Numerically, \(2\sqrt{10}\approx6.32\) and \(\sqrt{13}+\sqrt{17}\approx7.73\). Since \(6.32\le7.73\), the triangle inequality holds.

1. \(s=2+6i\); it is the diagonal vector from the parallelogram rule. 2. \(|z_1|=\sqrt{13}\), \(|z_2|=\sqrt{17}\), and \(|s|=2\sqrt{10}\) 3. \(2\sqrt{10}\le\sqrt{13}+\sqrt{17}\), so the triangle inequality is satisfied.

- Track how the real and imaginary parts change when you multiply by \(i\). - Recall the geometric meaning of multiplication by \(i\). - Compare the directions and lengths of the two adjacent side vectors.

1. \(w_1=i(5+2i)=5i+2i^2=-2+5i\). Multiplication by \(i\) rotates a vector \(90^\circ\) counterclockwise about the origin. 2. \(w_2=z+w_1=(5+2i)+(-2+5i)=3+7i\). 3. The vectors for \(z\) and \(w_1\) are perpendicular because one is a \(90^\circ\) rotation of the other. They also have equal magnitude because \(|i|=1\). A parallelogram with adjacent sides that are perpendicular and equal in length is a square.

1. \(w_1=-2+5i\); the transformation is a \(90^\circ\) counterclockwise rotation. 2. \(w_2=3+7i\) 3. The quadrilateral is a square.

- Use \(i^2=-1\) when multiplying. - Find distance by taking the magnitude of the difference. - Compare the locations of the two points relative to the origin. - How does multiplying the coordinates by \(i\) change the direction from the origin?

1. \(w=i(2+2i)=2i+2i^2=-2+2i\). 2. \(|z|=\sqrt{2^2+2^2}=2\sqrt{2}\), and \(|w|=\sqrt{(-2)^2+2^2}=2\sqrt{2}\). The magnitudes are equal. 3. The distance is \(|z-w|=|(2+2i)-(-2+2i)|=|4|=4\). 4. Multiplication by \(i\) is a \(90^\circ\) counterclockwise rotation about the origin. Because \(|i|=1\), the rotation does not change the distance from the origin.

a) \(w=-2+2i\) b) \(|z|=|w|=2\sqrt{2}\) c) \(4\) d) A \(90^\circ\) counterclockwise rotation about the origin

- Conjugation changes the sign of the imaginary part. - Equal complex numbers have equal real parts and equal imaginary parts. - A real sum has an imaginary part of zero.

1. Substitute \(z=a+bi\) and \(\overline{z}=a-bi\): \(2(a+bi)+(a-bi)=9+2i\). 2. Combine like terms to get \(3a+bi=9+2i\). 3. Equating real and imaginary parts gives \(3a=9\) and \(b=2\), so \(a=3\), \(b=2\), and \(z=3+2i\). 4. From \(z+w=12\), \(w=12-z=12-(3+2i)=9-2i\). 5. The point representing \(w\) is \((9, -2)\).

a) \(a=3\), \(b=2\), and \(z=3+2i\) b) \(w=9-2i\); the point is \((9, -2)\)

- Evaluate the expressions inside parentheses first. - Add real parts and imaginary parts separately. - Think of following several displacement vectors in sequence. - Use the associative property of real-number addition for each component.

1. \(w_1=(1+3)+(2-1)i=4+i\), and \(w_2=(3-2)+(-1+4)i=1+3i\). 2. \(S_1=(4+i)+(-2+4i)=2+5i\), and \(S_2=(1+2i)+(1+3i)=2+5i\). Therefore, \((z_1+z_2)+z_3=z_1+(z_2+z_3)\). 3. Grouping changes which intermediate vector sum is found first, but the total horizontal and vertical displacements are unchanged. Therefore, the endpoint is unchanged.

1. \(w_1=4+i\) and \(w_2=1+3i\) 2. \(S_1=S_2=2+5i\) 3. Grouping changes intermediate points but not the total displacement or final endpoint.

- Equal complex numbers have equal real parts and equal imaginary parts. - A real number has an imaginary part of zero. - Write an equation for the imaginary part of the difference.

1. \(z-w=(x-3)+(8-y)i\). Equating corresponding parts with \(2+5i\) gives \(x-3=2\) and \(8-y=5\). Therefore, \(x=5\) and \(y=3\). 2. For \(z-w\) to be real, its imaginary part must be zero. Thus, \(8-y=0\), so \(y=8\). The value of \(x\) may be any real number.

1. \(x=5\) and \(y=3\) 2. \(y=8\), with \(x\) arbitrary

- Equal complex numbers have matching real and imaginary parts. - A complex number lies on the imaginary axis when its real part is zero. - Write the difference in the form \(A+Bi\).

1. \(z_1+z_2=(a+3)+(2-b)i\). Equating parts with \(5-i\) gives \(a+3=5\) and \(2-b=-1\), so \(a=2\) and \(b=3\). 2. \(z_1-z_2=(a-3)+(2+b)i\). Its real part is zero when \(a-3=0\), so \(a=3\). The value of \(b\) may be any real number.

a) \(a=2\) and \(b=3\) b) \(a=3\), because the real part \(a-3\) must equal \(0\)

- Use the distributive property to expand each product. - Replace \(i^2\) with \(-1\). - The real part contains the terms without \(i\); the imaginary part is the real coefficient of \(i\).

1. Expand the first product: \((x+yi)(2-i)=2x-xi+2yi-yi^2=(2x+y)+(2y-x)i\). 2. Expand the second term: \(3(x-2yi)=3x-6yi\). 3. Subtract and combine corresponding parts: \(Z=(2x+y-3x)+(2y-x+6y)i=(-x+y)+(8y-x)i\). 4. Therefore, \(\operatorname{Re}(Z)=-x+y\) and \(\operatorname{Im}(Z)=8y-x\).

\(Z=(-x+y)+(8y-x)i\), \(\operatorname{Re}(Z)=-x+y\), and \(\operatorname{Im}(Z)=8y-x\)

- Multiply the binomials and use \(i^2=-1\). - Compare the point for \(z_1\) with the point for \(iz_1\). - Look for a multiplier that makes the imaginary terms cancel. - What special partner of \(z_1\) produces a real product?

1. \((3+2i)(-1+i)=-3+3i-2i+2i^2=-5+i\). 2. \(iz_1=i(3+2i)=-2+3i\). Multiplication by \(i\) rotates a point \(90^\circ\) counterclockwise about the origin. 3. Choose the conjugate \(c=3-2i\). Then \((3-2i)(3+2i)=3^2+2^2=13\), which is real. Any nonzero real multiple of \(3-2i\) also works.

1. \(w=-5+i\) 2. \(iz_1=-2+3i\); a \(90^\circ\) counterclockwise rotation 3. One choice is \(c=3-2i\)

- Equal complex numbers have matching real and imaginary parts. - First write the left side with a clear real part and imaginary part. - Solve the resulting system of two linear equations.

1. Substitute the expressions for \(u\) and \(v\): \((2x+yi)-(y-xi)=5+7i\). 2. Combine corresponding parts: \((2x-y)+(x+y)i=5+7i\). 3. Equating parts gives the system \(2x-y=5\) and \(x+y=7\). 4. Adding the equations gives \(3x=12\), so \(x=4\). Then \(4+y=7\), so \(y=3\).

\(x=4\) and \(y=3\)

- Track what happens when a factor containing \(i\) is squared. - Simplify both sides until the coefficients of \(i\) can be compared. - Substitute \(i^2=-1\) into the general expression. - Recall the sign of the square of a nonzero real number.

1. \((4i)(-2.5i)=-10i^2=10\). 2. \((xi)^2(2i)=x^2i^2(2i)=-2x^2i\). Thus, \(-2x^2i=-18i\), so \(x^2=9\) and \(x=\pm3\). 3. \(z^2=(ki)^2=k^2i^2=-k^2\). Since \(k\ne0\), \(k^2>0\), so \(-k^2<0\).

1) \(10\) 2) \(x\in\{-3, 3\}\) 3) \(z^2=-k^2<0\) for \(k\ne0\)

- Use the binomial-square formula. - Multiply by the conjugate when a complex number is in the denominator. - Replace \(i^2\) with \(-1\). - Combine real and imaginary parts separately at the end.

1. \((3-i)^2=9-6i+i^2=8-6i\). 2. Rationalize the denominator: \(\frac{20}{4+2i}\cdot\frac{4-2i}{4-2i}=\frac{80-40i}{16+4}=4-2i\). 3. Add the results: \((8-6i)+(4-2i)=12-8i\).

\(z=12-8i\)

- A real product has imaginary part zero. - A complex conjugate reverses the sign of the imaginary part. - Use the magnitude formula \(|a+bi|=\sqrt{a^2+b^2}\). - Calculate both sides of the magnitude identity separately.

1. \((k+2i)(3-i)=(3k+2)+(6-k)i\). The product is real when \(6-k=0\), so \(k=6\). 2. \(w\overline{w}=(a+bi)(a-bi)=a^2+b^2\). This is real and nonnegative because it is a sum of squares of real numbers. 3. \(z_1z_2=(1+2i)(2-i)=4+3i\), so \(|z_1z_2|=5\). Also, \(|z_1|=\sqrt{5}\) and \(|z_2|=\sqrt{5}\), so \(|z_1||z_2|=5\). The equality holds.

1) \(k=6\) 2) \(w\overline{w}=a^2+b^2\ge0\) 3) Yes; both sides equal \(5\)

- Use a common denominator for the two fractions. - Apply \(i^2=-1\). - Square both factors in \(2\sqrt{3}\). - Look for terms in the numerator that cancel.

1. Use the difference-of-squares pattern: \((2\sqrt{3})^2-(5i)^2=12-25\cdot(-1)=37\). 2. Use the common denominator \((x+yi)(x-yi)=x^2+y^2\). The numerator becomes \((x-yi)+(x+yi)=2x\). Therefore, the expression simplifies to \(\frac{2x}{x^2+y^2}\).

1) \(37\) 2) \(\frac{2x}{x^2+y^2}\)

- Isolate \(z\) by division. - Use the conjugate to simplify the complex denominator. - Apply the magnitude formula after finding rectangular form.

1. Divide by \(1-i\): \(z=\frac{3+5i}{1-i}\). 2. Multiply by the conjugate: \(z=\frac{(3+5i)(1+i)}{(1-i)(1+i)}\). 3. The numerator is \(-2+8i\), and the denominator is \(2\). Therefore, \(z=-1+4i\). 4. \(|z|=\sqrt{(-1)^2+4^2}=\sqrt{17}\).

a) \(z=-1+4i\) b) \(|z|=\sqrt{17}\)

- A real complex number has imaginary part zero. - First rewrite the quotient in rectangular form. - Track how \(k\) affects both components.

1. Multiply by the conjugate of the denominator: \(z=\frac{(k+4i)(1-2i)}{5}\). 2. Expanding gives \(z=\frac{k+8}{5}+\frac{4-2k}{5}i\). 3. For \(z\) to be real, \(4-2k=0\), so \(k=2\). 4. Then \(z=\frac{2+8}{5}=2\).

\(k=2\) and \(z=2\)

- Use \(X^2+Y^2=(X+iY)(X-iY)\). - Rewrite each expression as a sum of two squares. - Complete the square for the quadratic trinomial. - Use \(i^2=-1\) to check the product.

1. \(149=10^2+7^2\), so \(149=(10+7i)(10-7i)\). 2. \(25a^2+81b^2=(5a)^2+(9b)^2\), so it equals \((5a+9bi)(5a-9bi)\). 3. Complete the square: \(x^2-8x+25=(x-4)^2+3^2\). Therefore, it equals \((x-4+3i)(x-4-3i)\).

a) \((10+7i)(10-7i)\) b) \((5a+9bi)(5a-9bi)\) c) \((x-4+3i)(x-4-3i)\)

- Use the conjugate to separate the real and imaginary parts. - A number on the imaginary axis has real part zero. - Square the magnitude equation to remove the square root.

1. Multiply by the conjugate of the denominator: \(z=\frac{(k+3i)(3+i)}{10}\). 2. Expanding gives \(z=\frac{3k-3}{10}+\frac{k+9}{10}i\). 3. The real part is zero when \(3k-3=0\), so \(k=1\). 4. For \(|z|=\sqrt{2}\), square both sides: \(\frac{(3k-3)^2+(k+9)^2}{100}=2\). 5. Expanding gives \(10k^2+90=200\), so \(k^2=11\) and \(k=\pm\sqrt{11}\).

a) \(z=\frac{3k-3}{10}+\frac{k+9}{10}i\) b) \(k=1\) c) \(k=\pm\sqrt{11}\)

- A number on the imaginary axis has real part zero. - Use the conjugate to simplify the denominator. - Separate the real and imaginary parts before solving.

1. Multiply by the conjugate of the denominator: \(z=\frac{(a+6i)(2+i)}{5}\). 2. Expanding gives \(z=\frac{2a-6}{5}+\frac{a+12}{5}i\). 3. Set the real part equal to zero: \(2a-6=0\), so \(a=3\). 4. Substituting gives \(z=3i\).

\(a=3\) and \(z=3i\)

- A number on the imaginary axis has real part zero. - Square \(z\) and separate the real and imaginary parts. - Translate each condition into an equation or inequality. - Check which candidate makes the imaginary part positive.

1. \(z^2=(x+4i)^2=x^2-16+8xi\). 2. The real part is zero when \(x^2-16=0\), so \(x=4\) or \(x=-4\). 3. The imaginary part is \(8x\), which must be positive. Only \(x=4\) satisfies this condition.

\(x=4\)

- Factor a common power of \(i\) from the numerator and denominator. - Use exponent rules for division with the same base. - Cancel only nonzero common factors. - Check whether the simplified result still contains \(n\).

1. Factor the numerator: \(i^n+i^{n+1}=i^n(1+i)\). 2. Factor the denominator: \(i^{n+2}+i^{n+3}=i^{n+2}(1+i)\). 3. Since \(1+i\ne0\), cancel it to obtain \(w=\frac{i^n}{i^{n+2}}=i^{-2}\). 4. \(i^{-2}=\frac{1}{i^2}=-1\). The result no longer depends on \(n\).

\(w=-1\) for every \(n\in\mathbb{N}\)

- Try adding the first four powers of \(i\). - Find how the sum changes when a new group of four terms is added. - For part c, square the binomial once before multiplying again.

1. Four consecutive powers add to \(i+i^2+i^3+i^4=i-1-i+1=0\). 2. The first eight terms cancel in two groups of four. Thus \(S_{10}=i^9+i^{10}=i-1=-1+i\). 3. The partial sum is zero at the end of each complete four-term cycle and not after one, two, or three extra terms. Thus \(n=4,8,12,\ldots,100\), giving \(25\) values. 4. Since \((1+i)^2=1+2i+i^2=2i\), we have \((1+i)^4=(2i)^2=-4\), a real number.

a) \(S_{10}=-1+i\) b) \(n=4,8,12,\ldots,100\), a total of \(25\) values c) \((1+i)^4=-4\)

- Multiply the complex numbers in the numerator first. - Replace \(i^2\) with \(-1\). - Use the conjugate to make the denominator real.

1. Multiply the factors in the numerator: \((3+i)(1-2i)=5-5i\). 2. Multiply by the conjugate of the denominator: \(z=\frac{(5-5i)(2-i)}{(2+i)(2-i)}\). 3. The numerator is \(5-15i\), and the denominator is \(5\). Therefore, \(z=1-3i\).

\(1-3i\)

- Write the conjugate of \(a+bi\). - Simplify \(z\overline{z}\). - Equal complex numbers have matching real and imaginary parts. - Solve the resulting real equations.

1. Substitute \(z=a+bi\) and use \(z\overline{z}=a^2+b^2\). 2. The equation becomes \(a^2+b^2+2a+2bi=12+4i\). 3. Equating imaginary parts gives \(2b=4\), so \(b=2\). 4. Equating real parts gives \(a^2+4+2a=12\), or \(a^2+2a-8=0\). 5. Factoring gives \((a-2)(a+4)=0\), so \(a=2\) or \(a=-4\). 6. Therefore, \(z=2+2i\) or \(z=-4+2i\).

\(z\in\{2+2i, -4+2i\}\)

- Factor the base when helpful. - Use a binomial formula. - Simplify powers of \(i\). - For the fourth power, square the expression in stages.

1. \(2-2i=2(1-i)\). Since \((1-i)^2=-2i\), \((1-i)^4=(-2i)^2=-4\). Thus, \((2-2i)^4=2^4\cdot(-4)=-64\). 2. Expand the cube: \((-\sqrt{3}+i)^3=-3\sqrt{3}+9i+3\sqrt{3}-i=8i\).

1) \(-64\) 2) \(8i\)

- Apply the binomial-square formula. - Replace \(i^2\) with \(-1\). - A number on the imaginary axis has real part zero. - Separate the real and imaginary parts of \(z^2\).

1. \(z^2=[k+(k-1)i]^2=k^2-(k-1)^2+2k(k-1)i\). 2. Simplify the real part: \(k^2-(k-1)^2=2k-1\). 3. Set the real part equal to zero: \(2k-1=0\), so \(k=\frac{1}{2}\). 4. For this value, the imaginary coefficient is \(2\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)=-\frac{1}{2}\), so \(z^2=-\frac{1}{2}i\).

\(k=\frac{1}{2}\)

- Compare the expansions of a complex number and its conjugate. - Simplify the powers and reciprocals of \(i\) carefully. - For the claim about a reciprocal, use the defining property of multiplicative inverses.

1. \((a+bi)^2+(a-bi)^2=(a^2-b^2+2abi)+(a^2-b^2-2abi)=2(a^2-b^2)\), which is real. 2. For \(n=1\), \(i+i^{-1}=i-i=0\). For \(n=2\), \(i^2+i^{-2}=-1-1=-2\). 3. Oksana's claim is false. By definition, \(z\left(\frac{1}{z}\right)=1\) for every \(z\ne0\). This equals \(|z|\) only when \(|z|=1\). For example, if \(z=2\), the product is \(1\) but \(|z|=2\).

1. \(z^2+(\overline{z})^2=2(a^2-b^2)\), which is real. 2. \(0\) for \(n=1\); \(-2\) for \(n=2\) 3. The claim is false; \(z\cdot z^{-1}=1\), not generally \(|z|\).

- Use a conjugate to make a denominator real. - Compute \((1+i)^4\) by squaring twice. - A point on the imaginary axis has real part zero. - Track the signs when squaring a radical times \(i\).

1. \((1+i)^2=2i\), so \((1+i)^4=-4\). Then \(w=\frac{-4}{1-i}\cdot\frac{1+i}{1+i}=-2-2i\). 2. \(z^2=(1+\sqrt{3}i)^2=-2+2\sqrt{3}i\). Therefore, \(z^2-2z=(-2+2\sqrt{3}i)-2(1+\sqrt{3}i)=-4\), which is real. 3. \(\frac{x+i}{x-i}=\frac{(x+i)^2}{x^2+1}=\frac{x^2-1}{x^2+1}+\frac{2x}{x^2+1}i\). The real part is zero when \(x^2-1=0\), so \(x=\pm1\).

1. \(w=-2-2i\) 2. \(z^2=-2+2\sqrt{3}i\), and \(z^2-2z=-4\) 3. \(x=\pm1\)

- Rewrite each fraction with a real denominator. - Equal complex numbers have matching real and imaginary parts. - Put the equation in the form \(a+bi=c+di\). - Solve the resulting linear system.

1. Rationalize each denominator: \(\frac{x(1-i)}{2}+\frac{y(1+2i)}{5}=1\). 2. Multiply by \(10\): \(5x(1-i)+2y(1+2i)=10\). 3. Combine parts: \((5x+2y)+(4y-5x)i=10\). 4. Equating parts gives \(5x+2y=10\) and \(4y-5x=0\). 5. From \(5x=4y\), substitute into the first equation to obtain \(6y=10\), so \(y=\frac{5}{3}\). Then \(x=\frac{4}{3}\).

\(x=\frac{4}{3}\) and \(y=\frac{5}{3}\)

- A real complex number has imaginary part zero. - Use the conjugate to rewrite the quotient in rectangular form. - The denominator is always positive for real \(x\). - A fraction is zero when its numerator is zero.

1. Multiply by the conjugate of the denominator: \(w=\frac{(x+2i)(2-xi)}{(2+xi)(2-xi)}\). 2. The denominator is \(4+x^2\), and the numerator is \(4x+(4-x^2)i\). 3. The imaginary part is \(\frac{4-x^2}{4+x^2}\). Since the denominator is positive, this part is zero when \(4-x^2=0\). 4. Therefore, \(x=2\) or \(x=-2\).

\(x=\pm2\)

- Compare complex multiplication with multiplying two ordinary binomials. - Which products are missing if only the real parts and imaginary parts are multiplied separately? - Remember what happens when two factors of \(i\) are multiplied.

1. Complex multiplication follows the distributive property, so each term in the first factor multiplies each term in the second factor. 2. Expand: \((2+3i)(4-i)=8-2i+12i-3i^2\). 3. Since \(i^2=-1\), the last term is \(+3\), giving \(11+10i\). 4. Maya's rule omits the two cross-products and also fails to account correctly for \(i^2=-1\).

Maya's rule is invalid because complex multiplication requires distributing all terms. The correct product is \(11+10i\).

- Add and subtract a suitable \(x^2\)-term to create a difference of squares. - For part b, look for a sum-of-squares representation. - Create a perfect square of the form \((x^2\pm k)^2\).

1. For part a, add and subtract \(100x^2\): \(x^4+2500=(x^2+50)^2-(10x)^2\). 2. Factor the difference of squares: \(T=(x^2+10x+50)(x^2-10x+50)\). 3. For part b, rewrite \(T=(x^2-50)^2+(10x)^2\). 4. Therefore, \(T=(x^2-50+10xi)(x^2-50-10xi)\).

a) \((x^2+10x+50)(x^2-10x+50)\) b) \((x^2-50+10xi)(x^2-50-10xi)\)