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How do you convert #4-4i# to polar form?

1 Answer

A. S. Adikesavan

Apr 18, 2016

#4sqrt2 e^(i(7ou)/4)=4sqrt2(cos((7pi)/4)+i sin((7pi)/4))#

Explanation:

#x=4 and y=-4#.
#x+i y=r (cos theta + i sin theta)#

#r=sqrt(x^2+y^2)=4 sqrt 2#.

#cos theta=x/r=1/sqrt 2, sin theta=y/r=-1/sqrt2#

sine is negative and cosine is positive.
So, #theta# is in the fourth quadrant,
#theta=2pi-pi/4=(7pi)/4#.

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