Free math worksheets

- Under a power, treat the modulus and argument separately. - De Moivre’s theorem changes the argument by multiplying it by the exponent.

1. Square the modulus: \(2^2=4\). 2. Multiply the argument by \(2\): \(2\cdot\frac{\pi}{3}=\frac{2\pi}{3}\). 3. Therefore, \(z^2=4e^{i2\pi/3}\).

\(z^2=4e^{i2\pi/3}\)

- Find the distance from the origin and choose the angle in the stated interval. - Powers multiply the argument by the exponent. - A complex number is real when it lies on the horizontal axis. - Use the value of \(z^4\) to find \(z^8\).

1. \(|z|=\sqrt{2}\) and the argument in \([0,2\pi)\) is \(\frac{\pi}{4}\), so \(z=\sqrt{2}e^{i\pi/4}\). 2. \(z^n=(\sqrt{2})^ne^{in\pi/4}\) is real when \(\frac{n\pi}{4}\) is an integer multiple of \(\pi\). The smallest positive value is \(n=4\), and \(z^4=4e^{i\pi}=-4\). 3. \(z^8=(z^4)^2=16\), which is a positive integer.

1. \(z=\sqrt{2}e^{i\pi/4}\) 2. \(n=4\) and \(z^4=-4\) 3. \(z^8=16\)

- The fourth roots all have the same modulus. - Their arguments are equally spaced around a full circle. - Start from an argument of the original negative real number, then divide the complete family of arguments by \(4\).

1. Write \(-16=16e^{i\pi}\). The fourth roots have modulus \(2\) and arguments \(\frac{\pi+2\pi k}{4}\) for \(k=0,1,2,3\). 2. The four arguments are \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\). 3. Converting to rectangular form gives \(\sqrt{2}+i\sqrt{2}\), \(-\sqrt{2}+i\sqrt{2}\), \(-\sqrt{2}-i\sqrt{2}\), and \(\sqrt{2}-i\sqrt{2}\).

\(\left\{\sqrt{2}+i\sqrt{2},-\sqrt{2}+i\sqrt{2},-\sqrt{2}-i\sqrt{2},\sqrt{2}-i\sqrt{2}\right\}\)

- The root index determines how many distinct roots a nonzero complex number has. - Treat the modulus separately from the arguments. - Equal spacing divides one full revolution into equal parts. - Use the axis labels in the diagram to identify the root requested in part d.

1. A nonzero complex number has \(5\) distinct fifth roots. 2. If \(|z|^5=|32i|=32\), then \(|z|=\sqrt[5]{32}=2\). 3. The roots are equally spaced around a full circle, so the angular spacing is \(\frac{2\pi}{5}\). 4. The root on the positive imaginary axis is plotted at \((0, 2)\), and the diagram labels it \(B\).

a) \(5\) b) \(2\) c) \(\frac{2\pi}{5}\) d) Point \(B\)

- Find the modulus as the distance from the origin. - Use the quadrant to determine the argument. - Apply De Moivre’s theorem to powers. - A complex number is real when its sine component is zero.

1. The modulus is \(2\). Since the point lies in Quadrant II with reference angle \(\frac{\pi}{6}\), its argument is \(\frac{5\pi}{6}\). Thus, \(z=2e^{i5\pi/6}\). 2. \(z^6=2^6e^{i5\pi}=64(\cos5\pi+i\sin5\pi)=-64\). 3. \(z^n=2^ne^{i5n\pi/6}\) is real when \(\sin\left(\frac{5n\pi}{6}\right)=0\). Thus, \(\frac{5n}{6}\) must be an integer. Since \(5\) and \(6\) are relatively prime, \(n\) must be a positive multiple of \(6\).

1. \(z=2e^{i5\pi/6}\) 2. \(z^6=-64\) 3. \(n=6k\), where \(k\) is a positive integer

- An equation \(z^n=a\) has \(n\) complex roots. - The roots are equally spaced around a circle. - Find the roots in polar form, then convert to rectangular form. - Use conjugate factors when multiplying the nonreal roots.

1. Write \(8=8e^{i0}\). The cube roots are \(z_k=2e^{i2k\pi/3}\) for \(k=0, 1, 2\). 2. Therefore, \(z_0=2\), \(z_1=-1+i\sqrt{3}\), and \(z_2=-1-i\sqrt{3}\). 3. Their sum is \(2+(-1+i\sqrt{3})+(-1-i\sqrt{3})=0\). 4. Their product is \(2(-1+i\sqrt{3})(-1-i\sqrt{3})=2\cdot(1+3)=8\).

1. \(z_0=2\), \(z_1=-1+i\sqrt{3}\), \(z_2=-1-i\sqrt{3}\) 2. \(z_0+z_1+z_2=0\) 3. \(z_0z_1z_2=8\)

- Isolate \(z^3\) before converting the right side to polar form. - Use all coterminal arguments of the original complex number before dividing by the root index. - The three cube-root arguments should be equally spaced around the circle.

1. Rewrite the equation as \(z^3=-8i\). 2. In polar form with a radian argument in \([0,2\pi)\), \(-8i=8e^{i3\pi/2}\). 3. The cube roots have modulus \(2\) and arguments \(\frac{3\pi/2+2\pi k}{3}\) for \(k=0,1,2\). These are \(\frac{\pi}{2}\), \(\frac{7\pi}{6}\), and \(\frac{11\pi}{6}\). 4. Converting to rectangular form gives \(2i\), \(-\sqrt{3}-i\), and \(\sqrt{3}-i\).

\(z\in\{2i, -\sqrt{3}-i, \sqrt{3}-i\}\)

- Check whether the three reported angles actually determine three different directions. - Distinct roots come from different coterminal arguments of the original number before dividing the angle by the root index. - The arguments of three cube roots should be evenly spaced around the circle.

1. The three reported angles differ by whole multiples of \(2\pi\), so all three expressions represent the same complex number. Mateo added full turns after taking the cube root instead of using different arguments of \(8i\) before dividing by \(3\). 2. Write the arguments of \(8i\) as \(\frac{\pi}{2}+2\pi k\). Dividing by \(3\) gives root arguments \(\frac{\pi}{6}+\frac{2\pi k}{3}\). 3. For \(k=0,1,2\), the distinct arguments are \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), and \(\frac{3\pi}{2}\).

The reported roots are not distinct because their arguments differ by multiples of \(2\pi\). The three roots are \(2e^{i\pi/6}\), \(2e^{i5\pi/6}\), and \(2e^{i3\pi/2}\).

- Take the root of the modulus separately from the arguments. - Include the full family of coterminal arguments of the original complex number before dividing by the root index. - The three final root arguments should be equally spaced.

1. Each cube root has modulus \(\sqrt[3]{8}=2\). 2. The arguments satisfy \(3\theta=\frac{\pi}{4}+2\pi k\), so \(\theta=\frac{\pi}{12}+\frac{2\pi k}{3}\). 3. For \(k=0,1,2\), the arguments are \(\frac{\pi}{12}\), \(\frac{3\pi}{4}\), and \(\frac{17\pi}{12}\). 4. These three arguments lie in \([0,2\pi)\) and differ by \(2\pi/3\), so they give all three distinct roots.

\(z=2e^{i\pi/12},\ 2e^{i3\pi/4},\ 2e^{i17\pi/12}\)

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

- Translate “negative real” into a condition on the argument of the fourth power. - Work backward from the possible fourth-power arguments before using the quadrant condition. - The modulus of a fourth power is the fourth power of the original modulus.

1. Let the argument of \(z\) be \(\theta\). For \(z^4\) to be negative real, \(4\theta\equiv\pi\pmod{2\pi}\). 2. Thus \(\theta=\frac{\pi}{4}+\frac{k\pi}{2}\). The only such angle in Quadrant II is \(\theta=\frac{3\pi}{4}\). 3. Therefore, \(z=2\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)=-\sqrt{2}+i\sqrt{2}\). 4. Its fourth power has modulus \(2^4=16\) and argument \(4\cdot\frac{3\pi}{4}=3\pi\), so \(z^4=-16\).

\(z=-\sqrt{2}+i\sqrt{2}\) and \(z^4=-16\)