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Question Number 134419 by mohammad17 last updated on 03/Mar/21

$$whythefunctioncoszandsinzitis$$$$notboundedincomplexnumber?$$

Answered by Olaf last updated on 03/Mar/21

$$\cos z=\cos (x+iy)$$$$\cos z=\cos x\cos \left(iy\right)-\sin x\sin \left(iy\right)$$$$\cos z=\cos x\frac{e^{i\left(iy\right)}+e^{-i\left(iy\right)}}{2}-\sin x\frac{e^{i\left(iy\right)}-e^{-i\left(iy\right)}}{2i}$$$$\cos z=\cos x\frac{e^{-y}+e^y}{2}-\sin x\frac{e^{-y}-e^y}{2i}$$$$\cos z=\cos x\cosh y+i\sin x\sinh y$$$$\cos z\mathrm{is}\mathrm{not}\mathrm{bounded}\mathrm{because}$$$$\cosh y\mathrm{and}\sinh y\mathrm{are}\mathrm{not}\mathrm{bounded}.$$$$$$$$\mathrm{Same}\mathrm{demonstation}\mathrm{for}\sin z.$$$$$$

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