How can you use trigonometric functions to simplify # 19 e^( ( 3 pi)/4 i ) # into a non-exponential complex number?

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Answer 1 · 2017-10-29T13:35:30

# -(19root2 2) /2 + i(19root2 2)/2#

We need to consider how #re^(itheta) = rcostheta + irsintheta#

In this circumstance #theta# = #(3pi)/4#

And # r =19#

Hence #19e^((3ipi)/4)# = # 19( cos((3pi)/4) + isin((3pi)/4)) #

Hence via evaluating #cos((3pi)/4)# and #sin((3pi)/4)# we get;

# 19(-(root2 2) /2 + i(root2 2)/2)#

Hence yeilding our answer or # -(19root2 2) /2 + i(19root2 2)/2#