# Use Cauchy Riemann to prove this function is differentiable at all points

I expanded it out and got $e^{4z+1} = e^{4x+1}\cos{4y} + e^{4x+1}\sin{4y}$

Then my CR equations were -

$U_x = (4x+1)e^{4x+1}(4)(\cos 4y)$

$U_y = -e^{4x+1}(\sin4y)$

$V_x = (4x+1)e^{4x+1}(4)(\sin 4y)$

$V_y = e^{4x+1}(\cos 4y)$

Taking $U_x = V_y$ I get

$(4x+1)e^{4x+1}(4)(\cos 4y) = e^{4x+1}(\cos 4y)$

$(4)(4x+1) = 1$

But that can't be right as then it means the CR equations are only satisfied for a certain value of x. So what am I doing wrong?

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If you have $f(z) = f(x + iy)$ then a necessary and sufficient condition for it to be complex differentiable is that it satisfies the Cauchy-Riemann equations. This means that if you set $u(x,y) := \Re f$ and $v(x,y) := \Im f$ then you need to show that the following holds: $$u_x = v_y$$ and $$u_y = - v_x$$
In your case you have $$u(x,y) = \cos (4y) e^{4x + 1}$$ and $$v(x,y) = \sin (4y) e^{4x + 1}$$
This yields $$u_x = 4 \cos (4y) e^{4x + 1} = v_y$$ and $$u_y = -4 \sin (4y) e^{4x + 1} = - v_x$$
So you see that your $f$ is differentiable. Hope this helps.