Write $z _ { 1 } = 2 + 2 i$ in polar form then find $1 / z _ { 1 }$
A) $\frac { 1 } { 4 } - \frac { 1 } { 4 } i$

B) $\frac { 3 } { 4 } - \frac { 1 } { 4 } i$

C) $\frac { 1 } { 4 } + \frac { 1 } { 4 } i$

D) $- \frac { 1 } { 4 } - \frac { 1 } { 4 } i$

E) $\frac { 1 } { 4 } - \frac { \sqrt { 3 } } { 4 } i$

Question

Write $z _ { 1 } = 2 + 2 i$ in polar form then find $1 / z _ { 1 }$
A) $\frac { 1 } { 4 } - \frac { 1 } { 4 } i$

B) $\frac { 3 } { 4 } - \frac { 1 } { 4 } i$

C) $\frac { 1 } { 4 } + \frac { 1 } { 4 } i$

D) $- \frac { 1 } { 4 } - \frac { 1 } { 4 } i$

E) $\frac { 1 } { 4 } - \frac { \sqrt { 3 } } { 4 } i$

Quizplus · Accepted Answer

To convert $z_1 = 2 + 2i$ to polar form, we find the magnitude $r = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$ and the angle $\theta = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. Thus, $z_1$ in polar form is $2\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$. To find $1/z_1$, we use the polar form and get $1/(2\sqrt{2})(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))$, simplifying to $\frac{1}{4} - \frac{1}{4}i$.

Write z1=2+2iz _ { 1 } = 2 + 2 iz1​=2+2i

Write $z _ { 1 } = 2 + 2 i$