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

blestimd4pz

blestimd4pz

Answered question

2022-03-30

Prove that sin(z)=eizeiz2i=sin(x)cosh(y)+icos(x)sinh(y)
For complex z=x+iy. How to prove that
sin(z)=eizeiz2i=sin(x)cosh(y)+icos(x)sinh(y)
by using the power series definition
sin(z)=zz33!+z55!
and Euler's formula
ez=ex(cos(y)+isin(y))?

Answer & Explanation

Ali Price

Ali Price

Beginner2022-03-31Added 6 answers

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.
sinz=n=0(1)n(2n+1)!z2n+1=n=0i2n(2n+1)!z2n+1
=1in=0i2n+1(2n+1)!z2n+1=1in=0(iz)2n+1(2n+1)!
Consider 1(1)k2 which equals 1 for odd k and 0 for even k.
sinz=1ik=0(1(1)k2)(iz)kk!=12ik=01(1)kk!(iz)k
=12ik=0(iz)kk!12ik=0(iz)kk!
=eziezi2i
=e(x+yi)ie(x+yi)i2i
=ey+ξeyξ2i
=eyeξeyeξ2i
=ey(cosx+isinx)ey(cosxisinx)2i
=iisinxey+ey21icosxeyey2
=sinxcoshy+icosxsinhy

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