Free Math Worksheets

- Recall the repeating powers of \(i\). - Multiply the numerical factors first, then simplify the power of \(i\). - Track the signs of negative factors carefully. - Combine radicals before using \(i^2=-1\).

1. \(8\cdot0.5\,i^2=4\cdot(-1)=-4\). 2. \((-3)\cdot(-7)i^2=21\cdot(-1)=-21\). 3. \(\sqrt{2}\cdot\sqrt{18}\,i^2=\sqrt{36}\cdot(-1)=-6\). 4. \(24i^4=24(i^2)^2=24\).

1) \(-4\) 2) \(-21\) 3) \(-6\) 4) \(24\)

- 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) \(7\)

- 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 \(w=-3\)

- 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.

- 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: \((3a-4bi)-(5a+8bi)=-2a-12bi\).

\(-2a-12bi\)

- Multiply the binomials and use \(i^2=-1\). - Think about the location and argument of \(i\) in the complex plane. - Look for a multiplier that makes the imaginary terms cancel. - Recall the product of a complex number and its conjugate.

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 \(c=3-2i\)

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

- 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: \((8-6i)+(4-2i)=12-8i\).

\(z=12-8i\)

- 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 a conjugate to make the denominator real. - Use the difference-of-squares pattern. - Replace \(i^2\) with \(-1\).

1. Multiply by the conjugate of the denominator: \(z_1=\frac{(1+7i)(2+i)}{(2-i)(2+i)}\). The numerator is \(-5+15i\), and the denominator is \(5\), so \(z_1=-1+3i\). 2. \(z_2=\frac{10(3-i)}{(3+i)(3-i)}=\frac{10(3-i)}{10}=3-i\).

1) \(z_1=-1+3i\) 2) \(z_2=3-i\)

- 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: \((-5+12i)-(-5-12i)=24i\).

\(24i\)

- The powers of \(i\) repeat in a cycle. - Find each exponent modulo \(4\). - Add the simplified powers.

1. Since \(2023\equiv3\pmod{4}\), \(i^{2023}=-i\). 2. Since \(2024\equiv0\pmod{4}\), \(i^{2024}=1\). 3. Since \(2025\equiv1\pmod{4}\), \(i^{2025}=i\). 4. Therefore, \(z=-i+1+i=1\).

\(z=1\)

- Use the four-term cycle of powers of \(i\). - Separate powers that are multiples of \(4\). - Track the power of the negative sign in \((-i)^7\).

1. Since \(123\equiv3\pmod{4}\), \(i^{123}=-i\). 2. The four powers are \(-1\), \(-i\), \(1\), and \(i\), so their sum is \(0\). 3. \((-i)^7=(-1)^7i^7=i\), and \(i^{21}=i\). Their product is \(i^2=-1\).

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

- 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, \(z=1-3i\).

\(1-3i\)

- 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 to \(z\) and its conjugate. - Use the definition of a complex conjugate. - Simplify powers and reciprocals of \(i\). - Recall the defining property of a reciprocal.

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. The 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 \(|z|\).

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

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

- 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) \(z\overline{z}=a^2+b^2\ge0\)

- 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. - Recall the change in argument caused by multiplication by \(i\).

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 \(\operatorname{Im}(Z)=8y-x\)

- 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 \(k\ne0\)

- 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 \(n\in\mathbb{N}\)

- Compute several partial sums and look for a repeating pattern. - Determine how many consecutive powers sum to zero. - Square \(1+i\) first, then square the result.

1. Each block of four consecutive powers sums to zero: \(i+i^2+i^3+i^4=0\). Thus, \(S_{10}=i^9+i^{10}=-1+i\). 2. The sum is zero exactly at the end of each complete four-term cycle. Therefore, \(n\) must be a multiple of \(4\): \(n\in\{4, 8, 12, \ldots, 100\}\). 3. \((1+i)^2=2i\), so \((1+i)^4=(2i)^2=-4\), which is real.

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

- 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 \(z=-4+2i\).

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

- 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 \(z^2=-\frac{1}{2}i\).

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

- 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 \(x=-2\).

\(x=\pm2\)

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