Answer to Question #313820 in Complex Analysis for Jyo

Question #313820

Show that ๐‘ ๐‘–๐‘›๐‘–๐‘ฆ = ๐‘– ๐‘ ๐‘–๐‘›hy

1
Expert's answer
2022-03-20T06:43:32-0400

Solution in the first way. We apply the following formulas

sinโกz=eizโˆ’eโˆ’iz2i\sin z=\frac{e^{iz}-e^{-iz}}{2i} and sinhโกz=ezโˆ’eโˆ’z2\sinh z=\frac{e^{z}-e^{-z}}{2}. Then sinโกiy=ei(iy)โˆ’eโˆ’i(iy)2i=eโˆ’yโˆ’ey2i=โˆ’1ieyโˆ’eโˆ’y2=isinhโกy\sin iy=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=-\frac{1}{i}\frac{e^{y}-e^{-y}}{2}=i\sinh y, as required.


Solution in the second way. We apply the following formulas

sinโกz=โˆ‘n=0+โˆž(โˆ’1)n(2n+1)!z2n+1\sin z=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}z^{2n+1} and sinhโกz=โˆ‘n=0+โˆž1(2n+1)!z2n+1\sinh z=\sum\limits_{n=0}^{+\infty}\frac{1}{(2n+1)!}z^{2n+1}. Then

sinโกiy=โˆ‘n=0+โˆž(โˆ’1)n(2n+1)!(iy)2n+1=โˆ‘n=0+โˆž(โˆ’1)n(2n+1)!i2n+1y2n+1\sin iy=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}(iy)^{2n+1}=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}i^{2n+1}y^{2n+1}

=โˆ‘n=0+โˆž(โˆ’1)n(2n+1)!i(โˆ’1)ny2n+1=iโˆ‘n=0+โˆž1(2n+1)!y2n+1=isinhโกy=\sum\limits_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}i(-1)^ny^{2n+1}=i\sum\limits_{n=0}^{+\infty}\frac{1}{(2n+1)!}y^{2n+1}=i\sinh y, as required.


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