Free math worksheets

- Find the distance from the origin and the angle from the positive real axis. - 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 \(\arg(z)=\frac{\pi}{4}\), 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 \(\sqrt{2}-i\sqrt{2}\).

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

- 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 \(z^3\). - Write the right side in polar form. - Use the formula for complex roots. - The three roots are equally spaced in angle.

1. Rewrite the equation as \(z^3=-8i\). 2. In polar form, \(-8i=8(\cos270^\circ+i\sin270^\circ)\). 3. The cube roots have modulus \(2\) and arguments \(\frac{270^\circ+360^\circ k}{3}\) for \(k=0, 1, 2\). These arguments are \(90^\circ\), \(210^\circ\), and \(330^\circ\). 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 the different coterminal arguments of the original number before dividing the angle by the root index. - The arguments of three cube roots should be spaced evenly around the circle.

1. The three reported angles differ by whole multiples of \(2\pi\), so all three expressions represent the same complex number. The student added full turns after taking the cube root instead of using the 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 \(2e^{i3\pi/2}\).

- 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 \(z^4=-16\)