Let's find a conjugate to -5i+2
−5i+2=2+5i
let's find the rectangular representation of this number:
z=(2+5i)−1=2+5i1=(2+5i)(2−5i)2−5i=292−5i Real part
Re(z)=292
Imaginary part
Im(z)=−295 Modulus of z
∣z∣=2924+29225=2929 Re(z)>0 and Im(z)<0 hence
arg(z)=θ=2π−arctanRe(z)∣Im(z)∣=2π−arctan292295=2π−arctan25 Therefore the polar representation of z will be
z=2929(cos(2π−arctan25)+isin(2π−arctan25)) Exponential representation of z
z=2929ei(2π−arctan25) Geometric representation
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