How do you evaluate e^( ( 7 pi)/4 i) - e^( ( pi)/2 i)e7π4i−eπ2i using trigonometric functions?

1 Answer
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Jan 22, 2017

e^((7pi)/4i)-e^(pi/2i)=-1/sqrt(2)+(sqrt(2)+1)/sqrt(2)ie7π4i−eπ2i=−1√2+√2+1√2i

Explanation:

Euler's formula states

e^(ix)=cos(x)+isin(x)eix=cos(x)+isin(x)

Then for x=pi/2x=π2

e^(ix)=e^(x i)=e^(pi/2i)=cos(pi/2)+isin(pi/2)=0+i(1)=ieix=exi=eπ2i=cos(π2)+isin(π2)=0+i(1)=i

and for x=(7pi)/4=(2pi-pi/4) => -pi/4x=7π4=(2π−π4)⇒−π4

e^(-pi/4i)=cos(-pi/4)+isin(-pi/4)=cos(pi/4)-isin(pi/4)=1/sqrt(2)-1/sqrt(2)ie−π4i=cos(−π4)+isin(−π4)=cos(π4)−isin(π4)=1√2−1√2i

Then plug in

e^((7pi)/4i)-e^(pi/2i)=i-(1/sqrt(2)-1/sqrt(2)i)=i-1/sqrt(2)+1/sqrt(2)ie7π4i−eπ2i=i−(1√2−1√2i)=i−1√2+1√2i

=-1/sqrt(2)+(sqrt(2)+1)/sqrt(2)i=−1√2+√2+1√2i

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