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Question Number 142729 by mohammad17 last updated on 04/Jun/21

$$prove:\frac{d}{dz}\left({tan}^{-1}z\right)=\frac{1}{1+z^2}incomplexnumber$$$$helpmesir$$$$$$

Answered by mathmax by abdo last updated on 06/Jun/21

$$\mathrm{same}\mathrm{definition}\mathrm{of}\mathrm{differentiality}$$$$\frac{\mathrm{d}}{\mathrm{dz}}\left(\mathrm{arctanz}\right)={\lim }_{\mathrm{h}\to 0}\frac{\arctan (\mathrm{z}+\mathrm{h})-\mathrm{arctanz}}{\mathrm{h}}$$$${=}_{\mathrm{hospital}}={\lim }_{\mathrm{h}\to 0}\frac{1}{1+{(\mathrm{z}+\mathrm{h})}^2}=\frac{1}{1+{\mathrm{z}}^2}$$

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