Question #ecd7d

1 Answer
Douglas K.
Oct 17, 2017

The conversion to exponential form is:

#(a+bi)^n = (sqrt(a^2+b^2))^n e^(i ntan^-1(b/a))#

We know that #|e^(itheta)| = 1#

#:.#

#|(a+bi)^n| = (sqrt(a^2+b^2))^n#

Explanation:

Given: #(-1+i)^12#

we observe that #a = -1, b = 1, and n = 12#

#|(-1+i)^12| = (sqrt((-1)^2+1^2))^12#

#|(-1+i)^12| = 2^6#

#|(-1+i)^12| = 64#

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