Answer to Question #91339 in Complex Analysis for Ra

Question #91339
Describe the geometric, polar and exponential representations of:
(\bar -5i+2)⁻¹
1
Expert's answer
2019-07-05T08:54:14-0400

Let's find a conjugate to -5i+2


5i+2=2+5i\overline{-5i+2}=2+5i

let's find the rectangular representation of this number:


z=(2+5i)1=12+5i=25i(2+5i)(25i)=25i29z=( 2+5i )^{-1}= \frac{1}{2+5i}= \frac{2-5i}{(2+5i)(2-5i)}= \frac{2-5i}{29}

Real part


Re(z)=229Re(z)=\frac{2}{29}

Imaginary part


Im(z)=529Im(z)=-\frac{5}{29}

Modulus of z


z=4292+25292=2929|z|=\sqrt{\frac{4}{29^2}+\frac{25}{29^2}}=\frac{\sqrt{29}}{29}

Re(z)>0 and Im(z)<0 hence


arg(z)=θ=2πarctanIm(z)Re(z)=2πarctan529229=2πarctan52arg(z)=\theta=2\pi-arctan \frac{|Im(z)|}{Re(z)} ​ =2\pi-arctan \frac{\frac{5}{29}}{\frac{2}{29}}= 2\pi-arctan\frac{5}{2}

Therefore the polar representation of z will be


z=2929(cos(2πarctan52)+isin(2πarctan52))z=\frac{\sqrt{29}}{29}\left(cos\left(2\pi-arctan\frac{5}{2}\right )+isin\left(2\pi-arctan\frac{5}{2}\right )\right)

Exponential representation of z


z=2929ei(2πarctan52)z=\frac{\sqrt{29}}{29}e^{i\left(2\pi-arctan\frac{5}{2}\right )}

Geometric representation





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