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

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Answer 1 · 2017-01-22T22:16:27

#e^((7pi)/4i)-e^(pi/2i)=-1/sqrt(2)+(sqrt(2)+1)/sqrt(2)i#

Euler's formula states

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

Then for #x=pi/2#

#e^(ix)=e^(x i)=e^(pi/2i)=cos(pi/2)+isin(pi/2)=0+i(1)=i#

and for #x=(7pi)/4=(2pi-pi/4) => -pi/4#

#e^(-pi/4i)=cos(-pi/4)+isin(-pi/4)=cos(pi/4)-isin(pi/4)=1/sqrt(2)-1/sqrt(2)i#

Then plug in

#e^((7pi)/4i)-e^(pi/2i)=i-(1/sqrt(2)-1/sqrt(2)i)=i-1/sqrt(2)+1/sqrt(2)i#

#=-1/sqrt(2)+(sqrt(2)+1)/sqrt(2)i#