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Question Number 113346 by mohammad17 last updated on 12/Sep/20

Answered by mathmax by abdo last updated on 13/Sep/20

z =x+iy f(z) =cos(2z) =((e^(2iz) +e^(−2iz) )/2) =((e^(2i(x+iy)) +e^(−2i(x+iy)) )/2) =((e^(2ix−2y) +e^(−2ix) e^(2y) )/2) =((e^(−2y) {cos(2x)+isin(2x)}+e^(2y) {cos(2x)−isin(2x)})/2) =cos(2x).((e^(2y) +e^(−2y) )/2) −i sin(2x).((e^(2y) −e^(−2y) )/2) =cos(2x)ch(y)−i sin(2x)sh(2y) =u(x,y) +i v(x,y) with u(x,y) =cos(2x)ch2y and v(x,y) =−sin(2x)sh(2y) we have (∂u/∂x) =−2sin(2x)ch2y and (∂v/∂y) =−2sin(2x)ch(2y) ⇒ (∂u/∂x) =(∂v/∂y) (1) also (∂u/∂y) =2cos(2x)sh2y and −(∂v/∂x) =2cos(2x)sh(2y) ⇒ (∂u/∂y) =−(∂v/∂x) (2) so the condition of cauchy Rieman are verified

$$\mathrm{z}\:=\mathrm{x}+\mathrm{iy}\:\mathrm{f}\left(\mathrm{z}\right)\:=\mathrm{cos}\left(\mathrm{2z}\right)\:=\frac{\mathrm{e}^{\mathrm{2iz}} \:+\mathrm{e}^{−\mathrm{2iz}} }{\mathrm{2}}\:=\frac{\mathrm{e}^{\mathrm{2i}\left(\mathrm{x}+\mathrm{iy}\right)} \:+\mathrm{e}^{−\mathrm{2i}\left(\mathrm{x}+\mathrm{iy}\right)} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{2ix}−\mathrm{2y}} \:\:+\mathrm{e}^{−\mathrm{2ix}} \:\mathrm{e}^{\mathrm{2y}} }{\mathrm{2}}\:=\frac{\mathrm{e}^{−\mathrm{2y}} \left\{\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{isin}\left(\mathrm{2x}\right)\right\}+\mathrm{e}^{\mathrm{2y}} \left\{\mathrm{cos}\left(\mathrm{2x}\right)−\mathrm{isin}\left(\mathrm{2x}\right)\right\}}{\mathrm{2}} \\ $$$$=\mathrm{cos}\left(\mathrm{2x}\right).\frac{\mathrm{e}^{\mathrm{2y}} \:+\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}}\:\:−\mathrm{i}\:\:\mathrm{sin}\left(\mathrm{2x}\right).\frac{\mathrm{e}^{\mathrm{2y}} −\mathrm{e}^{−\mathrm{2y}} }{\mathrm{2}} \\ $$$$=\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{ch}\left(\mathrm{y}\right)−\mathrm{i}\:\mathrm{sin}\left(\mathrm{2x}\right)\mathrm{sh}\left(\mathrm{2y}\right)\:=\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)\:+\mathrm{i}\:\mathrm{v}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{with} \\ $$$$\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)\:=\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{ch2y}\:\mathrm{and}\:\mathrm{v}\left(\mathrm{x},\mathrm{y}\right)\:=−\mathrm{sin}\left(\mathrm{2x}\right)\mathrm{sh}\left(\mathrm{2y}\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\partial\mathrm{u}}{\partial\mathrm{x}}\:=−\mathrm{2sin}\left(\mathrm{2x}\right)\mathrm{ch2y}\:\:\:\mathrm{and}\:\frac{\partial\mathrm{v}}{\partial\mathrm{y}}\:=−\mathrm{2sin}\left(\mathrm{2x}\right)\mathrm{ch}\left(\mathrm{2y}\right)\:\Rightarrow \\ $$$$\frac{\partial\mathrm{u}}{\partial\mathrm{x}}\:=\frac{\partial\mathrm{v}}{\partial\mathrm{y}}\:\left(\mathrm{1}\right)\:\mathrm{also}\:\frac{\partial\mathrm{u}}{\partial\mathrm{y}}\:=\mathrm{2cos}\left(\mathrm{2x}\right)\mathrm{sh2y}\:\:\:\mathrm{and}\:−\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\:=\mathrm{2cos}\left(\mathrm{2x}\right)\mathrm{sh}\left(\mathrm{2y}\right)\:\Rightarrow \\ $$$$\frac{\partial\mathrm{u}}{\partial\mathrm{y}}\:=−\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\:\:\left(\mathrm{2}\right)\:\:\mathrm{so}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{of}\:\mathrm{cauchy}\:\mathrm{Rieman}\:\mathrm{are}\:\mathrm{verified} \\ $$

Commented by mohammad17 last updated on 13/Sep/20

thank you sir can you help me in question 4

$${thank}\:{you}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{question}\:\mathrm{4} \\ $$

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