What is the trigonometric form of #5e^( (3pi)/2 i ) #?

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Answer 1 · 2018-03-05T14:13:39

See the explanation below

Apply Euler's Identity

#e^(itheta)=costheta+isintheta#

Therefore,

The trigononetric form is

#z=5e^(3/2pii)=5(cos(3/2pi)+isin(3/2pi))#

#=5(0-i)#

#=-5i#

And upon simplification

#z=-5i#