Use the definition e^(itheta)=costheta+isinthetaeiθ=cosθ+isinθ
e^((5ipi)/4)=cos((5pi)/4)+isin((5pi)/4)e5iπ4=cos(5π4)+isin(5π4)
e^((7ipi)/4)=cos((7pi)/4)+isin((7pi)/4)e7iπ4=cos(7π4)+isin(7π4)
cos((5pi)/4)=-sqrt2/2cos(5π4)=−√22
cos((7pi)/4)=sqrt2/2cos(7π4)=√22
sin((7pi)/4)=-sqrt2/2sin(7π4)=−√22
sin((5pi)/4)=-sqrt2/2sin(5π4)=−√22
The result is
e^((5ipi)/4)-e^((7ipi)/4)=cos((5pi)/4)+isin((5pi)/4)-(cos((7pi)/4)+isin((7pi)/4))e5iπ4−e7iπ4=cos(5π4)+isin(5π4)−(cos(7π4)+isin(7π4))
e^((5ipi)/4)-e^((7ipi)/4)=-sqrt2/2-isqrt2/2-sqrt2/2+isqrt2/2e5iπ4−e7iπ4=−√22−i√22−√22+i√22
So the final answer is
e^((5ipi)/4)-e^((7ipi)/4)=-sqrt2e5iπ4−e7iπ4=−√2
