How do you convert #-1+i to polar form?

1 Answer
Ratnaker Mehta
Jun 15, 2016

#(r,theta)=(sqrt2,3pi/4).#

Explanation:

To convert a complex no. #z=x+iy# into polar form, we need to find #r# & #theta# such that #x=rcostheta, y=rsintheta#, where #theta in (-pi,pi]# and #r=sqrt(x^2+y^2).#

In our case, #x=-1, y=1,# so, clearly, #r=sqrt2.#

#x=rcostheta rArr -1=sqrt2costheta rArr costheta=-1/sqrt2, (-ve)#
#y=rsintheta rArr 1=sqrt2sintheta rArr sintheta=1/sqrt2, (+ve)#

We conclude that #theta# lies in the #II^(nd)# Quadrant, #theta=3pi/4.#

Thus #(r,theta)=(sqrt2,3pi/4)#is the desired polar form of the complex no. #z=-1+i.#

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