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Question Number 115968 by Engr_Jidda last updated on 29/Sep/20

Answered by 1549442205PVT last updated on 30/Sep/20

$$\mathrm{Put}\mathrm{u}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2-1;\mathrm{v}(\mathrm{x},\mathrm{y})={\mathrm{y}}^2+1$$$$\mathrm{The}\mathrm{condition}\mathrm{Cauchy}-\mathrm{Rieman}\mathrm{for}$$$$\mathrm{the}\mathrm{analytical}\mathrm{is}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}}\left(1\right),\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{x}}\left(2\right)$$$$\mathrm{We}\mathrm{have}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}=2\mathrm{x},\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=0,\frac{\partial \mathrm{v}}{\partial \mathrm{x}}=0,\frac{\partial \mathrm{v}}{\partial \mathrm{y}}=2\mathrm{y}$$$$\mathrm{Henece},\mathrm{we}\mathrm{need}\mathrm{must}\mathrm{have}$$$$2\mathrm{x}=2\mathrm{y}(\mathrm{second}\mathrm{condition}\mathrm{is}\mathrm{satisfied}$$$$\mathrm{\forall }\mathrm{x},\mathrm{y})\iff \mathrm{y}=\mathrm{x}$$$$\mathrm{Thus},\mathrm{f}\left(\mathrm{z}\right)\mathrm{is}\mathrm{analytical}\mathrm{function}$$$$\mathrm{on}\mathrm{the}\mathrm{line}\mathrm{y}=\mathrm{x}\mathrm{and}\mathrm{we}\mathrm{have}:$$$$\mathrm{f}{}^{\prime }\left(\mathrm{z}\right)=\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{i}\frac{\partial \mathrm{v}}{\partial \mathrm{x}}=\frac{\partial \mathrm{u}}{\partial \mathrm{x}}-\mathrm{i}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}}-\mathrm{i}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}}+\mathrm{i}\frac{\partial \mathrm{v}}{\partial \mathrm{x}}$$$$=2\mathrm{x}=2\mathrm{y}$$

Commented by Engr_Jidda last updated on 30/Sep/20

$$\frac{\partial f}{\partial z}=\frac{1}{2}(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y})$$$$=\frac{1}{2}(2x-i\left(2iy\right))=x+y$$$$thankyousimuchsir$$

Commented by Engr_Jidda last updated on 30/Sep/20

$$thankyouverymuchsir$$

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