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. \(z^8=16\)

- 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 \(\sqrt{3}-i\).

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