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Question Number 145822 by Mrsof last updated on 08/Jul/21

$$findzerosfunctionf\left(z\right)=1-cosz$$

Commented by Mrsof last updated on 08/Jul/21

$$???????$$

Answered by mathmax by abdo last updated on 09/Jul/21

$${\mathrm{z}}_{\mathrm{k}}=2\mathrm{k}\pi \mathrm{are}\mathrm{roots}$$$$\mathrm{f}\left(\mathrm{z}\right)=0\Rightarrow 1-\mathrm{cosz}=0\Rightarrow \mathrm{cosz}=1\Rightarrow \frac{{\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}}{2}=1\Rightarrow {\mathrm{e}}^{\mathrm{iz}}+{\mathrm{e}}^{-\mathrm{iz}}-2=0$$$$\Rightarrow \mathrm{u}+{\mathrm{u}}^{-1}-2=0(\mathrm{u}={\mathrm{e}}^{\mathrm{iz}})\Rightarrow {\mathrm{u}}^2+1-2\mathrm{u}=0\Rightarrow {(\mathrm{u}-1)}^2=0\Rightarrow \mathrm{u}=1$$$${\mathrm{e}}^{\mathrm{iz}}=1={\mathrm{e}}^{\mathrm{i}\left(2\mathrm{k}\pi \right)}\Rightarrow {\mathrm{z}}_{\mathrm{k}}=2\mathrm{k}\pi (\mathrm{k}\in \mathrm{Z})\mathrm{are}\mathrm{roots}\mathrm{of}\mathrm{f}$$

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