Question

Buna! Cum se determina partea reala pentru (2+3i)/(3-2i)?

Answer 1

(2+3i)/(3-2i)
(2+3i)(3+2i)/9-4i^2
6+4i+9i+6i^2/9+4
6-6+13i/13
=i.=> partea reala=0.

Answer 2

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