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What is the trigonometric form of #9 e^( (3pi)/2 i ) #?

1 Answer

Nov 27, 2017

#9*(cos((3pi)/2)+isin((3pi)/2))#

Explanation:

Recall Eulers formula:
#e^(iz)=cosz+isinz#

In this case #z=(3pi)/2#

Therefore #9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))#

If you wish to simplify this further, #cos((3pi)/2)=0#, and #sin((3pi)/2)=-1#, therefore:

#9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))=9*(0+i*(-1))=-9i#

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