How do you use De Moivre's Theorem to simplify #z=(2[cos(5pi/4)+isin(5pi/4)])^5# to #16sqrt2 + 16sqrt2 i#? Trigonometry The Polar System De Moivre’s and the nth Root Theorems 1 Answer P dilip_k Dec 10, 2016 #z=(2[cos(5pi/4)+isin(5pi/4)])^5# #=>z=2^5[cos((5xx5pi)/4)+isin((5xx5pi)/4)]# #=>z=2^5[cos(6pi+pi/4)+isin(6pi+pi/4)]# #=>z=2^5[cos(pi/4)+isin(pi/4)]# #=>z=32[1/sqrt2+i/sqrt2]# #=>z=16sqrt2+16sqrt2i# Answer link Related questions How do you use De Moivre’s Theorem to find the powers of complex numbers in polar form? What is the DeMoivre's theorem used for? How do you find the #n^{th}# roots of complex numbers in polar form? How do you find #[2(\cos 120^\circ + i \sin 120^\circ)]^5# using the De Moivre's theorem? What is #(-\frac{1}{2}+\frac{i\sqrt{3}}{2})^{10}#? How do you find the three cube roots of #-2-2i \sqrt{3}#? If the roots can be determined, will some form of De Moivre’s Theorem be used? How do you find the two square roots of 2i? Question #f6317 How do you find #z, z^2, z^3, z^4# given #z=sqrt2/2(1+i)#? See all questions in De Moivre’s and the nth Root Theorems Impact of this question 3388 views around the world You can reuse this answer Creative Commons License