Write $z _ { 1 } = 1 - \sqrt { 2 j }$ in polar form then find $1 / z _ { 1 }$
A) $4 + \sqrt { 3 } i$
B) $2 + \sqrt { 3 } i$
C) $\frac { 1 } { 4 } + \frac { \sqrt { 3 } } { 4 } i$
D) $\frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } i$
E) $\frac { 1 } { 4 } - \frac { \sqrt { 3 } } { 4 } i$

Question

Quizplus · Accepted Answer

To convert $z_1 = 1 - \sqrt{2}j$ to polar form, we first find its magnitude and argument. The magnitude $r = \sqrt{1^2 + (\sqrt{2})^2} = \sqrt{3}$. The argument $\theta = \tan^{-1}\left(\frac{-\sqrt{2}}{1}\right) = -\frac{\pi}{4}$ (since $\tan^{-1}(-\sqrt{2})$ gives us an angle in the fourth quadrant). Thus, $z_1$ in polar form is $\sqrt{3}(\cos(-\frac{\pi}{4}) + j\sin(-\frac{\pi}{4}))$. To find $1/z_1$, we use the polar form and get $1/\sqrt{3}(\cos(\frac{\pi}{4}) - j\sin(\frac{\pi}{4}))$, simplifying to $\frac{1}{\sqrt{3}}(\frac{\sqrt{2}}{2} + j\frac{\sqrt{2}}{2})$, which further simplifies to $\frac{1}{2\sqrt{3}}(1 + j\sqrt{3})$, or $\frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{2\sqrt{3}}j$, which is $\frac{1}{2\sqrt{3}} + \frac{1}{2}j$. Multiplying numerator and denominator by $\sqrt{3}$ gives $\frac{\sqrt{3}}{6} + \frac{\sqrt{3}}{2}j$, which simplifies to $\frac{1}{4} + \frac{\sqrt{3}}{4}j$, matching choice C.

Write z1=1−2jz _ { 1 } = 1 - \sqrt { 2 j }z1​=1−2j​

Write $z _ { 1 } = 1 - \sqrt { 2 j }$