Question

Partea reala a numarului complex z=2-i/2+i

Answer 1

Răspuns:

3/5

Explicație pas cu pas:

$z=\frac{2-i}{2+i}=\frac{(2-i)(2-i)}{(2+i)(2-i)}=\frac{4-4i+i^{2} }{2^{2}-i^{2}}=\frac{4-4i-1}{4-(-1)}=\frac{3-4i}{5}=\frac{3}{5}-\frac{4}{5}i , deci\\Re(z)=\frac{3}{5}$

Answer 2

z=(2-i)/(2+i)=(2-i)²/(2-i)(2+i)=(4+i²-4i)/(4-i²)=(4-1-4i)/(4+1)=(-4i+3)/5=-4i/3+3/5

Re(z)=3/5