How do you simplify # [(3+2i)^ 3 / (-2+3i)^4] #?

1 Answer
Cesareo R.
Jun 27, 2016

# (3-2i)/13#

Explanation:

# 3+2i=sqrt(3^2+2^2)e^{i phi}#
#-2+3i=sqrt(3^2+2^2)e^{i( phi+pi/2)}#

with #phi = arctan(2/3)#

# [(3+2i)^ 3 / (-2+3i)^4] =((3+2i)/(-2+3i))^3/( (-2+3i))=e^{-i (3pi)/2}/(sqrt(3^2+2^2)e^{i( phi+pi/2)})#
#=1/sqrt(3^2+2^2)e^{-i(phi+2pi)} = 1/sqrt(3^2+2^2)e^{-i phi} =#
#sqrt(3^2+2^2)/(3^2+2^2)e^{-i phi} = (3-2i)/13#

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