imaginary number evaluation

Question

Question Let $z_1 = 1 + i$, $z_2 = 2 - i$, evaluate $$\left | \frac{z_1}{z_2} \right |$$

I have this question! Its to evaluate the fraction !

what I did is the following

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

I got $\frac{3i+1}{5}$ simplest form

and then I tried doing it the other way around !

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

and I got $\frac{2}{(1-3i)}$

I don't know if there is another way to solve it

Answer 1

The first part is the right way to go(where you got $\displaystyle \frac{3i+1}{5}$). After that, you can calculate the absolute value of a complex number a+bi as $\sqrt{a^2+b^2}$.

Answer 2

$|z_1/z_2|=|(1+3i)/5|$ is correct. Then, note that $|a+bi|=\sqrt{a^2+b^2}$ where $a,b\in\mathbb R.$

Answer 3

Youre meant to be evaluating the modulus of your answer, so $|\frac{3i+1}{5}| = \sqrt{\frac{9}{25}+\frac{1}{25}} = \sqrt{\frac{10}{25}} = \frac{\sqrt{10}}{5} = \frac{\sqrt{2}}{\sqrt{5}} = B$

Answer 4

You do not have to evaluate the complex fraction: Simply use $|z_1| = \sqrt{1^2+1^2} =\sqrt{2}\;$ and $|z_2|=\sqrt{2^2+1^2}=\sqrt{5}\;$ and the fact $$\left| \frac{z_1}{z_2}\right| =\frac{|z_1|}{|z_2|} =\frac{\sqrt{2}}{\sqrt{5}}$$

Answer 5

You have: $$R=\left|\dfrac{z_1}{z_2}\right|=\left|\dfrac{1+i}{2-i}\right|=\left|\dfrac{(1+i)(2+i)}{(2-i)(2+i)}\right|=\left|\dfrac{1+3i}{5}\right|$$ So: $$R=\dfrac{\sqrt{10}}{5}=\dfrac{\sqrt{2}}{\sqrt{5}}$$