Free Math Worksheets

- Look for the repeating cycle in the powers of \(i\). - Recall how to write the square root of a negative number using \(i\). - Use the quadratic formula.

1. The powers of \(i\) are \(i^1=i\), \(i^2=-1\), \(i^3=-i\), and \(i^4=1\). 2. Therefore, \(S=i-1-i+1=0\). 3. For \(x^2+4x+13=0\), the discriminant is \(4^2-4\cdot1\cdot13=-36\). 4. Since \(\sqrt{-36}=6i\), the quadratic formula gives \(x=\frac{-4\pm6i}{2}=-2\pm3i\).

a) \(S=0\) b) \(\{-2+3i, -2-3i\}\)

- Determine when the imaginary part of \(z^2\) is zero. - Expand \((a+bi)^2\) and identify its two components. - A negative real square suggests a number on the imaginary axis. - Use \(i^2=-1\).

1. \(z^2=(a+bi)^2=a^2-b^2+2abi\). 2. The square is real exactly when its imaginary part is zero: \(2ab=0\). Thus, \(a=0\) or \(b=0\). 3. To solve \(z^2=-16\), the case \(b=0\) would require \(a^2=-16\), which has no real value of \(a\). 4. In the case \(a=0\), \(-b^2=-16\), so \(b=\pm4\). 5. Therefore, \(z=4i\) or \(z=-4i\).

a) \(a=0\) or \(b=0\) b) \(z\in\{4i, -4i\}\)

- Begin with \((z-z_1)(z-z_2)=0\). - Use \(i^2=-1\) when multiplying the roots. - Relate the sum and product of the roots to the coefficients.

1. Use the factored equation \((z-z_1)(z-z_2)=0\). Thus, \(b=-(z_1+z_2)\) and \(c=z_1z_2\). 2. \(z_1+z_2=(2+i)+(1-3i)=3-2i\), so \(b=-3+2i\). 3. \(z_1z_2=(2+i)(1-3i)=2-6i+i-3i^2=5-5i\). 4. The equation is \(z^2+(-3+2i)z+(5-5i)=0\).

\(b=-3+2i\), \(c=5-5i\); the equation is \(z^2+(-3+2i)z+(5-5i)=0\)

- Recall the conjugate-root theorem for polynomials with real coefficients. - Use the sum and product of the roots. - A negative discriminant gives nonreal roots.

1. Nonreal roots of a polynomial with real coefficients occur in conjugate pairs, so \(z_2=3+4i\). 2. By the relationships between roots and coefficients, \(p=-(z_1+z_2)=-6\). 3. Also, \(q=z_1z_2=(3-4i)(3+4i)=25\). 4. The discriminant is \(D=(-6)^2-4\cdot25=-64\). Since \(D<0\), the equation has no real solutions.

a) \(z_2=3+4i\) b) \(p=-6\) and \(q=25\) c) \(D=-64\), so there are no real solutions

- Use the quadratic formula with a negative discriminant. - Recall the relationships between roots and coefficients. - Interpret the roots as points in the complex plane. - Use a base and perpendicular height to find the triangle area.

1. The quadratic formula gives \(z=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm6i}{2}=2\pm3i\). 2. The sum is \((2+3i)+(2-3i)=4\). 3. The product is \((2+3i)(2-3i)=4+9=13\). 4. The triangle has vertices \((0, 0)\), \((2, 3)\), and \((2, -3)\). Its vertical base has length \(6\), and the perpendicular distance from the origin to the line \(x=2\) is \(2\). 5. The area is \(\frac{1}{2}\cdot6\cdot2=6\).

a) \(z_1=2+3i\) and \(z_2=2-3i\) b) \(z_1+z_2=4\) and \(z_1z_2=13\) c) \(6\) square units

- Expand the square and separate real and imaginary parts. - Set up two real equations by matching components. - Use one equation to eliminate a variable. - Reject values that cannot be squares of real numbers.

1. Expand: \((x+yi)^2=x^2-y^2+2xyi\). 2. Equating parts gives \(x^2-y^2=3\) and \(xy=2\). 3. Since \(x\ne0\), \(y=\frac{2}{x}\). Substitution gives \(x^2-\frac{4}{x^2}=3\). 4. Multiply by \(x^2\): \(x^4-3x^2-4=0\). Let \(u=x^2\), so \(u^2-3u-4=0\). 5. The values are \(u=4\) or \(u=-1\). Because \(u=x^2\ge0\), \(x^2=4\), so \(x=\pm2\). 6. Using \(xy=2\), the corresponding values are \(y=1\) when \(x=2\), and \(y=-1\) when \(x=-2\).

\((x, y)=(2, 1)\) or \((x, y)=(-2, -1)\)