Neka je z=3−2iz=3-2iz=3−2i.
a) Izračunati z⋅zˉz\cdot\bar zz⋅zˉ, zˉz\dfrac{\bar z}{z}zzˉ i 1z\dfrac{1}{z}z1.
b) Naći realni i imaginarni deo broja w=4+7i2+i+1−3i1+iw=\dfrac{4+7i}{2+i}+\dfrac{1-3i}{1+i}w=2+i4+7i+1+i1−3i.
a) zzˉ=∣z∣2=32+(−2)2=13z\bar z=|z|^2=3^2+(-2)^2=13zzˉ=∣z∣2=32+(−2)2=13. Dalje
zˉz=3+2i3−2i=(3+2i)213=5+12i13,1z=zˉ∣z∣2=3+2i13.\frac{\bar z}{z}=\frac{3+2i}{3-2i}=\frac{(3+2i)^2}{13}=\frac{5+12i}{13},\qquad \frac{1}{z}=\frac{\bar z}{|z|^2}=\frac{3+2i}{13}.zzˉ=3−2i3+2i=13(3+2i)2=135+12i,z1=∣z∣2zˉ=133+2i.
b) 4+7i2+i=(4+7i)(2−i)5=15+10i5=3+2i\dfrac{4+7i}{2+i}=\dfrac{(4+7i)(2-i)}{5}=\dfrac{15+10i}{5}=3+2i2+i4+7i=5(4+7i)(2−i)=515+10i=3+2i; 1−3i1+i=(1−3i)(1−i)2=−2−4i2=−1−2i\dfrac{1-3i}{1+i}=\dfrac{(1-3i)(1-i)}{2}=\dfrac{-2-4i}{2}=-1-2i1+i1−3i=2(1−3i)(1−i)=2−2−4i=−1−2i.
Zbir: w=(3+2i)+(−1−2i)=2w=(3+2i)+(-1-2i)=2w=(3+2i)+(−1−2i)=2, pa je Rew=2\operatorname{Re}w=2Rew=2, Imw=0\operatorname{Im}w=0Imw=0.
Izvor: Do indeksa (autorski)