Respuesta :
so-called simplifying, really means, "rationalizing the denominator", which is another way of saying, "getting rid of that pesky radical in the bottom"
[tex]\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[/tex]
[tex]\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[/tex]
Multiply by the conjugate (flip the symbol in the denominator and multiply numerator and denominator by it):
[tex]\frac{4\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}[/tex]
Distribute on top and bottom; for bottom:
[tex]3^2+\left(-2\right)^2=13[/tex]
For top:
[tex]4\cdot \:3+4\cdot \:2i=12+8i[/tex]
Now we have:
[tex]\frac{12+8i}{13}[/tex]
Final answer:
[tex]\frac{12}{13}+\frac{8}{13}i[/tex]
[tex]\frac{4\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}[/tex]
Distribute on top and bottom; for bottom:
[tex]3^2+\left(-2\right)^2=13[/tex]
For top:
[tex]4\cdot \:3+4\cdot \:2i=12+8i[/tex]
Now we have:
[tex]\frac{12+8i}{13}[/tex]
Final answer:
[tex]\frac{12}{13}+\frac{8}{13}i[/tex]