Let's find a conjugate to -5i+2
− 5 i + 2 ‾ = 2 + 5 i \overline{-5i+2}=2+5i − 5 i + 2 = 2 + 5 i
let's find the rectangular representation of this number:
z = ( 2 + 5 i ) − 1 = 1 2 + 5 i = 2 − 5 i ( 2 + 5 i ) ( 2 − 5 i ) = 2 − 5 i 29 z=( 2+5i )^{-1}= \frac{1}{2+5i}= \frac{2-5i}{(2+5i)(2-5i)}= \frac{2-5i}{29} z = ( 2 + 5 i ) − 1 = 2 + 5 i 1 = ( 2 + 5 i ) ( 2 − 5 i ) 2 − 5 i = 29 2 − 5 i Real part
R e ( z ) = 2 29 Re(z)=\frac{2}{29} R e ( z ) = 29 2
Imaginary part
I m ( z ) = − 5 29 Im(z)=-\frac{5}{29} I m ( z ) = − 29 5 Modulus of z
∣ z ∣ = 4 2 9 2 + 25 2 9 2 = 29 29 |z|=\sqrt{\frac{4}{29^2}+\frac{25}{29^2}}=\frac{\sqrt{29}}{29} ∣ z ∣ = 2 9 2 4 + 2 9 2 25 = 29 29 Re(z)>0 and Im(z)<0 hence
a r g ( z ) = θ = 2 π − a r c t a n ∣ I m ( z ) ∣ R e ( z ) = 2 π − a r c t a n 5 29 2 29 = 2 π − a r c t a n 5 2 arg(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} a r g ( z ) = θ = 2 π − a rc t an R e ( z ) ∣ I m ( z ) ∣ = 2 π − a rc t an 29 2 29 5 = 2 π − a rc t an 2 5 Therefore the polar representation of z will be
z = 29 29 ( c o s ( 2 π − a r c t a n 5 2 ) + i s i n ( 2 π − a r c t a n 5 2 ) ) 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) z = 29 29 ( cos ( 2 π − a rc t an 2 5 ) + i s in ( 2 π − a rc t an 2 5 ) ) Exponential representation of z
z = 29 29 e i ( 2 π − a r c t a n 5 2 ) z=\frac{\sqrt{29}}{29}e^{i\left(2\pi-arctan\frac{5}{2}\right )} z = 29 29 e i ( 2 π − a rc t an 2 5 ) Geometric representation
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