IIT JEE , AIEEE Forum - Mathematics , Algebra

neil_mahaseth

z= i - 1/(cos

/3 + i sin

/3)

srujana

$eq=z= i - rac{1}{(cos rac{pi}{3} + i sin rac{pi}{3})}$

$eq=z= i - {(cos rac{pi}{3} - i sin rac{pi}{3})}$

$eq=\sqrt{ -(cos^2 \frac{\pi}{3}+(1+sin \frac{\pi}{3})^2)}[ \frac{cos \frac{\pi}{3}}{\sqrt{cos^2 \frac{\pi}{3}+(1+sin \frac{\pi}{3})^2}}-i \frac{sin \frac{\pi}{3}+1}{\sqrt{cos^2 \frac{\pi}{3}+(1+sin \frac{\pi}{3})^2}}]$

$eq=which is of the form z=x(cos heta -isin heta)$

$eq=x =-(\sqrt{{cos^2 \frac{\pi}{3}+(i(1+sin \frac{\pi}{3})^2}})$

$eq=cos \theta = \frac{cos \frac{\pi}{3}}{\sqrt{cos^2 \frac{\pi}{3}+(i(1+sin \frac{\pi}{3})^2)}}$

$eq= sin \theta = \frac{(1+sin \frac{\pi}{3})}{\sqrt{cos^2 \frac{\pi}{3}+(i(1+sin \frac{\pi}{3})^2)}}$

bsgdabest

its not clear if its " (i-1)/dr" or " i - (dr)^-1"

if its the 1st one, write nr as sqrt 2* cis (3pi/4) and dr as cis(pi/3) therefore your expression becomes z= sqrt2*(cis (5pi/12) since (cisA)/(cisB)= cis(A-B)
here in polar coordiantes, your r is distance from pole which is modulus of the complex no. and the "5pi/12" is the angle. therfore the polar coordinate must be
(

2,5pi/12) you can try the same thing for the 2 nd case.

bsgdabest

i- cis(-pi/3)
= -0.5 + i(i +

3 /2 )
a+ib can be written in euler form as r.e^{i( arc tan b/a)}and r²= a^2 + b^2 .

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