Question

convert the following in the polar form

1+3i/1-2i


Pls answer quick

Answer 1

this is your answer
hope it helps : )

Answer 2

Answer:

$\sqrt{2}e^{i\frac{3\pi}{4}}$

Step-by-step explanation:

The given equation is:

${\frac{1+3i}{1-2i}$

We have to convert the given equation in the polar form, thus

=${\frac{1+3i}{1-2i}{\times}\frac{1+2i}{1+2i}}$

=${\frac{1+5i-6}{1+4}}$

=${\frac{-5+5i}{5}}$

=$-1+i$

=${\sqrt{2}{\times}{\frac{1}{\sqrt{2}}(-1+i)$

=$\sqrt{2}(\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$

=$\sqrt{2}(cos({\pi}-{\frac{\pi}{4})+isin({\pi}-{\frac{\pi}{4}))$

=$\sqrt{2}(cos{\frac{3\pi}{4}+isin{\frac{3\pi}{4})$

=$\sqrt{2}e^{i\frac{3\pi}{4}}$

which is the required polar form.