Question

100 POINTS AND BRAINLIEST FOR THIS WHOLE SEGMENT

a) Find zw, Write your answer in both polar form with ∈ [0, 2pi] and in complex form.

b) Find z^10. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.

c) Find z/w. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.

d) Find the three cube roots of z in complex form. Give answers correct to 4 decimal

places.

Answer 1

Answer:

See Below (Boxed Solutions).

Step-by-step explanation:

We are given the two complex numbers:

$\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)$

First, convert z to polar form. Recall that polar form of a complex number is:

$z=r\left(\cos \theta + i\sin\theta\right)$

We will first find its modulus r, which is given by:

$\displaystyle r = |z| = \sqrt{a^2+b^2}$

In this case, a = √3 and b = -1. Thus, the modulus is:

$r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2$

Next, find the argument θ in [0, 2π). Recall that:

$\displaystyle \tan \theta = \frac{b}{a}$

Therefore:

$\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}$

Evaluate:

$\displaystyle \theta = -\frac{\pi}{6}$

Since z must be in QIV, using reference angles, the argument will be:

$\displaystyle \theta = \frac{11\pi}{6}$

Therefore, z in polar form is:

$\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)$

Part A)

Recall that when multiplying two complex numbers z and w:

$zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)$

Therefore:

$\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)$

Simplify. Hence, our polar form is:

$\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}$

To find the complex form, evaluate:

$\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}$

Part B)

Recall that when raising a complex number to an exponent n:

$\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)$

Therefore:

$\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)$

Substitute:

$\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)$

Simplify:

$\displaystyle z^{10} = 1024\left(\cos\frac{55\pi}{3}+i\sin \frac{55\pi}{3}\right)$

Simplify using coterminal angles. Thus, the polar form is:

$\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}$

And the complex form is:

$\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}$

Part C)

Recall that:

$\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)$

Therefore:

$\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)$

Simplify. Hence, our polar form is:

$\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}$

And the complex form is:

$\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}$

Part D)

Let a be a cube root of z. Then by definition:

$\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)$

From the property in Part B, we know that:

$\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)$

Therefore:

$\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)$

If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:

$\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}$

The first equation can be easily solved:

$r=\sqrt[3]{2}$

For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:

$\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}$

Solve for the argument:

$\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}$

There are three distinct solutions within [0, 2π):

$\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}$

Hence, the three roots are:

$\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)$

Or, approximately:

$\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}$