Prove that sin⁡(z)=eiz−e−iz2i=sin⁡(x)cosh(y)+icos⁡(x)sinh(y)For complex z=x+iy. How to prove thatsin⁡(z)=eiz−e−iz2i=sin⁡(x)cosh(y)+icos⁡(x)sinh(y)by using the power series definitionsin⁡(z)=z−z33!+z55!−…and Euler's formulaez=ex(cos⁡(y)+isin⁡(y))?

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Prove that sin⁡(z)=eiz−e−iz2i=sin⁡(x)cosh(y)+icos⁡(x)sinh(y)For complex z=x+iy. How to prove thatsin⁡(z)=eiz−e−iz2i=sin⁡(x)cosh(y)+icos⁡(x)sinh(y)by using the power series definitionsin⁡(z)=z−z33!+z55!−…and Euler&#039;s formulaez=ex(cos⁡(y)+isin⁡(y))?

Ali Price · Accepted Answer

Is it really necessary to use the power series definition of sine? The left equality is easier to prove right-to-left with the exp. power expansion.sin⁡z=∑n=0∞(−1)n(2n+1)!z2n+1=∑n=0∞i2n(2n+1)!z2n+1=1i∑n=0∞i2n+1(2n+1)!z2n+1=1i∑n=0∞(iz)2n+1(2n+1)!Consider 1−(−1)k2  which equals 1 for odd k and 0 for even k.sin⁡z=1i∑k=0∞(1−(−1)k2)(iz)kk!=12i∑k=0∞1−(−1)kk!(iz)k=12i∑k=0∞(iz)kk!−12i∑k=0∞(−iz)kk!=ezi−e−zi2i=e(x+yi)i−e−(x+yi)i2i=e−y+ξ−ey−ξ2i=e−yeξ−eye−ξ2i=e−y(cos⁡x+isin⁡x)−ey(cos⁡x−isin⁡x)2i=ii⋅sin⁡x⋅ey+e−y2−1i⋅cos⁡x⋅ey−e−y2=sin⁡xcoshy+icos⁡xsinhy

Prove that \(\displaystyle{\sin{{\left({z}\right)}}}={\frac{{{e}^{{{i}{z}}}-{e}^{{-{i}{z}}}}}{{{2}{i}}}}={\sin{{\left({x}\right)}}}{\text{cosh}{{\left({y}\right)}}}+{i}{\cos{{\left({x}\right)}}}{\text{sinh}{{\left({y}\right)}}}\) For complex z=x+iy. How to

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