Let O be the origin and A be the point z 1 = 1 + 2 i . If B is the point z 2 , Re z 2 < 0 , such that O…

Given $O$ be the origin and $A$ be the point $z_1=1+2i$,

$B$ is the point $z_2,Re\left(z_2\right)<0$,

Also $OAB$ is a right angled isosceles triangle with $OB$ as hypotenuse,

Plotting the diagram we get, 

 

Now by using concept of rotation we get,

$\frac{z_2-0}{\left(1+2i\right)-0}=\frac{\left|OB\right|}{\left|OA\right|}{e}^{\frac{i\pi }{4}}$

$\Rightarrow \frac{z_2}{1+2i}=\sqrt{2}{e}^{\frac{i\pi }{4}}$

$Or z_2=\left(1+2i\right)\left(1+i\right)$

$=-1+3i$

Now finding $\arg z_2=\pi -{\tan }^{-1}\frac{\left|3\right|}{\left|1\right|}=\mathrm{\pi }-{\tan }^{-1}3$

Now $\left|z_2\right|=\sqrt{1^2+3^2}=\sqrt{10}$

Now solving $z_1-2z_2=\left(1+2i\right)+2-6i=3-4i$

So, $\arg \left(z_1-2z_2\right)=-{\tan }^{-1}\frac{4}{3}$

Now solving $\left|2z_1-z_2\right|=\left|2+4i+1-3i\right|=\left|3+i\right|$

$=\sqrt{10}$

So, $\left|2z_1-z_2\right|=5$ is not true.