Question

Determinati partea imaginara a nr z=2+3i supra 3-2i. Detaliat

Answer 1

$z= \frac{2+3i}{3-2i}= \frac{(2+3i)(3+2i)}{(3-2i)(3+2i)}= \frac{6+4i+9i+6i^2}{9-4i^2} = \frac{13i}{13}=i \\ z=0+1i \\ Rez=0 ;Imz=1$