How do you evaluate # e^( ( pi)/4 i) - e^( ( 35 pi)/4 i)# using trigonometric functions?

Question

1315 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-28T09:18:34

#e^((pi)/4i)-e^((35pi)/4i)=sqrt2#

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

#e^((pi)/4i)=cos((pi)/4)+isin((pi)/4)=1/sqrt2+i1/sqrt2#

#e^((35pi)/4i)=cos((35pi)/4)+isin((35pi)/4)#

= #cos(8pi+(3pi)/4)+isin(8pi+(3pi)/4)#

= #cos((3pi)/4)+isin((3pi)/4)#

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

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

= #-1/sqrt2+i1/sqrt2#

Hence #e^((pi)/4i)-e^((35pi)/4i)=(1/sqrt2+i1/sqrt2)-(-1/sqrt2+i1/sqrt2)#

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

= #2/sqrt2+0=sqrt2#