Question

The real part of $\frac{1+2i}{1-3i}$ is $a$. Find $2a+1$.

Answer 1

$\Rightarrow \frac{1+2i}{1-3i}=\frac{(1+2i)(1+3i)}{(1-3i)(1+3i)}=\frac{1+5i-6}{1+9}=\frac{-5+5i}{10}=\frac{-1}{2}+\frac{1}{2}$

$=a+ib$

$a=-\frac{1}{2}$

$2a=-1$

$2a+1=0$