Question

Simplify the given expression below 4/3-2i

Answer 1

So-called simplifying, really means, "rationalizing the denominator", which is another way of saying, "getting rid of that pesky radical in the bottom"


$\bf \cfrac{4}{3-2i}\cdot \cfrac{3+2i}{3+2i}\impliedby \textit{multiplying by the conjugate of the bottom} \\\\\\ \cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{4(3+2i)}{3^2-(2i)^2}\implies \cfrac{4(3+2i)}{3^2-(4i^2)}\\\\ -------------------------------\\\\ recall\qquad i^2=-1\\\\ -------------------------------\\\\ \cfrac{4(3+2i)}{3^2-(4\cdot -1)}\implies \cfrac{4(3+2i)}{9+4}\implies \cfrac{12+8i}{13}\implies \cfrac{12}{13}+\cfrac{8}{13}i$

Answer 2

Multiply by the conjugate (flip the symbol in the denominator and multiply numerator and denominator by it): 
$\frac{4\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}$

Distribute on top and bottom; for bottom: 
$3^2+\left(-2\right)^2=13$

For top: 
$4\cdot \:3+4\cdot \:2i=12+8i$

Now we have:
$\frac{12+8i}{13}$

Final answer: 
$\frac{12}{13}+\frac{8}{13}i$