Question

Express $\frac{{(1+i)}^3}{4+3i}$ in the from a+ib

Answer 1

We have,

$\frac{{(1+i)}^3}{4+3i}$

$\frac{1^3+i^3+3i(1+i)}{4+3i}$

$\Rightarrow \frac{1-i+3i+3i^2}{4+3i}$

$\Rightarrow \frac{1-i+3i-3}{4+3i}$

$\Rightarrow \frac{-2+2i}{4+3i}$

Now, Rationalize and we get,

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

$\Rightarrow \frac{-8-6i+8i-6i^2}{4^2-{\left(3i\right)}^2}$

$\Rightarrow \frac{-8+2i+6}{16+9}$

$\Rightarrow \frac{2i-2}{25}$

$\Rightarrow \frac{2i}{25}-\frac{2}{25}$

Hence, this is the answer.