Question

Write the complex number $\frac{(2-3i)}{(3+4i)}$ in the form $A+iB$.

Answer 1

$\frac{(2-3i)}{(3+4i)}$

Multiplying both numerator and denominator by the Complex conjugate of the denominator

Therefore,

$\frac{(2-3i)}{(3+4i)}=\frac{2-3i}{3+4i}\times \frac{3-4i}{3-4i}$

$=\frac{(2-3i)(3-4i)}{{\left(3\right)}^2-{\left(4i\right)}^2}$

$=\frac{(6+12i^2)-(9+8)i}{9+16}$

$=\frac{-6-17i}{25}$

$=\frac{-6}{25}+\frac{-17}{25}i$

Here, $A=\frac{-6}{25}$ and $B=\frac{-17}{25}$