Question

Calculati partea imaginara a numarului
z= 2-i supra 1 +i

Answer 1

$\it z = \dfrac{^{1-i)}2-i}{\ 1+i}= \dfrac{2-i-2i+i^2}{1^2-i^2}= \dfrac{2-3i-1}{1-(-1)} = \dfrac{1-3i}{1+1} = \dfrac{1-3i}{2}=\\ \\ \\ = \dfrac{1}{2}- \dfrac{3}{2}i \Longrightarrow Imz= -\dfrac{3}{2}$

Answer 2

z=2-i /1+i=(2-i)·(1-i) /2=2-2i-i-1 /2=1-3i /2 => z=1-3i /2=1/2 -3i/2 => partea imaginara a numarului complex z este Im(z)=-3/2 .