How can you use trigonometric functions to simplify # 7 e^( ( 3 pi)/4 i ) # into a non-exponential complex number?

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Answer 1 · 2016-11-20T17:47:39

The answer is #=-(7sqrt2)/2+(i7sqrt2)/2#

We know that #e^(itheta)=costheta+isintheta#

So, #7e^(3pii/4)=7(cos ((3pi)/4)+isin((3pi)/4))#

#cos(3pi/4)=-sqrt2/2#

#sin(3pi/4)=sqrt2/2#

Therefore,

#7e^(3pii/4)=7(-sqrt2/2+isqrt2/2)#

#-(7sqrt2)/2+(i7sqrt2)/2#