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

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Answer 1 · 2016-03-27T09:23:17

#e^((pi)/4i)-e^((7pi)/4i)=sqrt2i#

As #e^(itheta)=costheta+isintheta#, we have

#e^((pi)/4i)=cos((pi)/4)+isin((pi)/4)# and

#e^((7pi)/4i)=cos((7pi)/4)+isin((7pi)/4)#

Hence, #e^((pi)/4i)-e^((7pi)/4i)#

= #(cos((pi)/4)+isin((pi)/4))-(cos((7pi)/4)+isin((7pi)/4))#

As #cos(pi/4)=cos((7pi)/4)=1/sqrt2#,

#sin(pi/4)=1/sqrt2# and #sin((7pi)/4)=-1/sqrt2#

#e^((pi)/4i)-e^((7pi)/4i)#

= #(1/sqrt2+i1/sqrt2)-(1/sqrt2+i(-1/sqrt2))#

= #(1/sqrt2-1/sqrt2+i(1/sqrt2)+i(1/sqrt2))#

=#0+i2/sqrt2# = #sqrt2i#