How do you convert #-6+6i# to polar form?

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Answer 1 · 2016-09-30T08:09:29

# -6+6i=6sqrt 2(cos(3/4pi) +i sin (3/4pi))#

The general #theta inQ_2# is #2kpi+3/4pi, k=0, +-1, |=2, +-3, ...#

The conversion formula for z=x+iy=(x, y) is

#(x, y)=r(cos theta, sin theta)#, giving

#r = sqrt(x^2+y^2)# (the principal square root) #>=0#,

#cos theta =x/r and sin theta = y/r#

Here, #x = -6 and y = 6#. So,

#rsqrt((-6)^2+6^2)=6sqrt 2#

#cos theta =-6/(6sqrt 2)=-1/sqrt 2 and sin theta=6/(6sqrt 2)=1/sqrt 2#.

Note that there is no common principal value for both.

In #(0, 2pi)#, the value is #3/4piin Q_2#., wherein

sine is positive and cosine is negative..

And so,

# -6+6i=6sqrt 2(cos(3/4pi) +i sin (3/4pi))#

The general #theta inQ_2# is #2kpi+3/4pi, k=0, +-1, |=2, +-3, ...#