How do you change #10z^12# into polar coordinates where #z = 2 +4i#? Precalculus Complex Numbers in Trigonometric Form Trigonometric Form of Complex Numbers 1 Answer Tom Dec 28, 2015 #2+4i# #theta = arctan(b/a)# #theta = arctan(2)# #r = sqrt(2^2+4^2) = sqrt(20) = 2sqrt(5)# polar form of #2+4i##=>##2sqrt(5)(cos(arctan(2))+isin(arctan(2)))# #(2sqrt(5)(cos(arctan(2))+isin(arctan(2))))^12# using the moivre formula #(2sqrt(5))^12(cos(arctan(2))+isin(arctan(2)))^12# #6,4*10^7(cos(12arctan(2))+isin(12arctan(2))) = z^12# #6,4*10^8(cos(12arctan(2))+isin(12arctan(2))) = 10z^12# Answer link Related questions How do I find the trigonometric form of the complex number #-1-isqrt3#? How do I find the trigonometric form of the complex number #3i#? How do I find the trigonometric form of the complex number #3-3sqrt3 i#? How do I find the trigonometric form of the complex number #sqrt3 -i#? How do I find the trigonometric form of the complex number #3-4i#? How do I convert the polar coordinates #3(cos 210^circ +i\ sin 210^circ)# into rectangular form? What is the modulus of the complex number #z=3+3i#? What is DeMoivre's theorem? How do you find a trigonometric form of a complex number? Why do you need to find the trigonometric form of a complex number? See all questions in Trigonometric Form of Complex Numbers Impact of this question 1469 views around the world You can reuse this answer Creative Commons License