## Related questions with answers

Write each complex number in polar form. Then use either (4) or (5) to obtain a polar form of the given number. Write the polar form in the form $a+i b$. $\frac{-i}{2-2 i}$

Solution

VerifiedLet us first write $z_1=-i$ and $z_2=2-2i$ in polar form before proceeding. Since $z_1$ lies on the negative part of the imaginary axis it is easy to see that its polar form is

$z_1 = |i| \left( \cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right) \right) =\cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right)$

The modulus of $z_2$ is

$r_2 = |2-2i| = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = 2\sqrt{2} .$

Since $z_2$ lies in the fourth quadrant we know that $\arg z_2=\theta$ where $\theta$ is the solution to

$\tan \theta = \frac{-2}{2} = -1 \Rightarrow \theta = - \frac{\pi}{4} .$

We conclude that the polar form of $z_2$ is

$z_2 = 2\sqrt{2} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)$

Finally, we apply (5) and get

$\begin{aligned} \frac{-i}{2-2i} & = \frac{\cos \left( - \frac{\pi}{2} \right) + i \sin \left( - \frac{\pi}{2} \right)}{2\sqrt{2} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)} = \\ & = \frac{1}{2\sqrt{2}} \left( \cos \left( - \frac{\pi}{2} - \left( - \frac{\pi}{4} \right) \right) + i \sin \left( - \frac{\pi}{2} - \left( - \frac{\pi}{4} \right) \right) \right) = \\ & = \pmb{\frac{\sqrt{2}}{4} \left( \cos \left( - \frac{\pi}{4} \right) + i \sin \left( - \frac{\pi}{4} \right) \right)} = \\ & = \frac{\sqrt{2}}{4} \left( \frac{\sqrt{2}}{2} - i \sin \frac{\sqrt{2}}{2} \right) = \pmb{\frac{1}{4} - \frac{1}{4} i }\end{aligned}$

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