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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
	$z =x+iy f(z) =cos(2z) =((e^(2iz) +e^(−2iz) )/2) =((e^(2i(x+iy)) +e^(−2i(x+iy)) )/2) =((e^(2ix−2y) +e^(−2ix) e^(2y) )/2) =((e^(−2y) {cos(2x)+isin(2x)}+e^(2y) {cos(2x)−isin(2x)})/2) =cos(2x).((e^(2y) +e^(−2y) )/2) −i sin(2x).((e^(2y) −e^(−2y) )/2) =cos(2x)ch(y)−i sin(2x)sh(2y) =u(x,y) +i v(x,y) with u(x,y) =cos(2x)ch2y and v(x,y) =−sin(2x)sh(2y) we have (∂u/∂x) =−2sin(2x)ch2y and (∂v/∂y) =−2sin(2x)ch(2y) ⇒ (∂u/∂x) =(∂v/∂y) (1) also (∂u/∂y) =2cos(2x)sh2y and −(∂v/∂x) =2cos(2x)sh(2y) ⇒ (∂u/∂y) =−(∂v/∂x) (2) so the condition of cauchy Rieman are verified$
	$z = x + iy f (z) = \cos (2 z) = \frac{e^{2 iz} + e^{- 2 iz}}{2} = \frac{e^{2 i (x + iy)} + e^{- 2 i (x + iy)}}{2}$ $= \frac{e^{2 ix - 2 y} + e^{- 2 ix} e^{2 y}}{2} = \frac{e^{- 2 y} {\cos (2 x) + isin (2 x)} + e^{2 y} {\cos (2 x) - isin (2 x)}}{2}$ $= \cos (2 x) . \frac{e^{2 y} + e^{- 2 y}}{2} - i \sin (2 x) . \frac{e^{2 y} - e^{- 2 y}}{2}$ $= \cos (2 x) ch (y) - i \sin (2 x) sh (2 y) = u (x, y) + i v (x, y) with$ $u (x, y) = \cos (2 x) ch 2 y and v (x, y) = - \sin (2 x) sh (2 y) we have$ $\frac{\partial u}{\partial x} = - 2 \sin (2 x) ch 2 y and \frac{\partial v}{\partial y} = - 2 \sin (2 x) ch (2 y) \Rightarrow$ $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} (1) also \frac{\partial u}{\partial y} = 2 \cos (2 x) sh 2 y and - \frac{\partial v}{\partial x} = 2 \cos (2 x) sh (2 y) \Rightarrow$ $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} (2) so the condition of cauchy Rieman are verified$
	Commented by mohammad17 last updated on 13/Sep/20

	$t h a n k y o u s i r c a n y o u h e l p m e i n q u e s t i o n 4$
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