How do you simplify #[(1+i)^4 / (2-2i)^3]#?

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Answer 1 · 2016-08-02T09:00:18

#(1-i)/8#

#(1+i)^4 / (2-2i)^3=1/2^3(1+i)^4/(1-i)^3=1/2^3(1+i)^7/((1-i)^3(1+i)^3) = (1+i)^7/2^6#

Now using de Moivre's identity

#1+i = sqrt(2)e^{ipi/4}# so

#(1+i)^7 = (sqrt(2))^7e^{i7/4pi}# but

#e^{i7/4pi} = e^{i 8/4pi} e^{-ipi/4} = e^{-ipi/4}#

Finally

#(1+i)^4 / (2-2i)^3=(2^3 sqrt(2)e^{-ipi/4})/2^6=sqrt(2)e^{-ipi/4}/2^3 = (1-i)/8#