Question

find modulus and argument of 1+2i/1-3i

Answer 1

We have,,

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

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

$\sf{=\frac{(1+2i)(1+3i)}{1-9i^{2}}}$

$\sf{=\frac{1+3i+2i+6i^{2}}{1+9}}$

$\sf{=\frac{1+5i-6}{10}}$

$\sf{=\frac{-5+5i}{10}}$

$\sf{=\frac{-5}{10}+\frac{5i}{10}}$

$\sf{=\frac{-1}{2}+\frac{i}{2}}$

Here, x = -1/2 and y = 1/2

Therefore,

Modulus, $\sf{r = \sqrt{x^{2}+y^{2}}}$

$\implies{\sf{r = \sqrt{(\frac{-1}{2})^{2}+(\frac{1}{2})^{2}}}}$

$\implies{\sf{r = \sqrt{(\frac{2}{4})}}}$

$\implies{\boxed{\sf{r =\frac{1}{\sqrt{2}}}}}$

Now,

$\sf{\tan(\Theta)=|\frac{y}{x}|}$

$\implies{\sf{\tan(\Theta)=|\frac{\frac{1}{2}}{\frac{-1}{2}}|}}$

$\implies{\sf{\tan(\Theta)=1}}$

therefore,

Argument, $\boxed{\sf{\Theta=\frac{\pi}{4}}}$

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