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I need to find where $f(x+iy)=-6(\cos x+i\sin x)+(2-2i)y^3+15(y^2+2y)$ is complex differentiable.

I first rearranged the function into its real and imaginary parts: $f(x+iy)=(-6\cos x+2y^3+15y^2+30y)+i(-6\sin x-6y^2)$

That means $u(x,y)=-6\cos x+2y^3+15y^2+30y$ and $v(x,y)=-6\sin x-6y^3$.

Then, if we take the partial derivative of u and v in terms of x and y:

$u_x=6\sin x$

$u_y=6y^2+30y+30$

$v_x=-6\cos x$

$v_y=-18y^2$

Then, by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$.

This means that: $6\sin x=-18y^2$ and $6y^2+30y+30=6\cos x$.

This is where I am stuck. How do I solve for x and y? I was thinking that I could proceed in this way:

$\sin^2 x + \cos^2 x=1 \Rightarrow (-3y^2)^2+(y^2+5y+5)^2=1 \Rightarrow 10y^4+10y^3+35y^2+50y+24=0$

However, from here, how do I solve for y and then solve for x? I'd appreciate any tips. Thanks for your help in advance!

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2 Answers

up vote 1 down vote accepted

You can see from wolfram alpha that the solutions for the pair of equations never overlap.

But using your chain of logic: $10y^4+10y^3+35y^2+50y+24=0$ has no real solutions. Therefore the function is complex-differentiable nowhere.

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We can simplify these equations into $$ \sin{x} = -3 y^2, \quad y^2 + 5y + 5 = \cos{x} $$ Now, $x$ and $y$ are real, so the only way that the first equation can be satisfied is if $y \in [-1/\sqrt{3}, 1/\sqrt{3}]$, since otherwise $-3y^2$ will not be in the range of sine.

Now we turn to the second equation. The graph of $y^2 + 5y + 5$ achieves its minimum at $y = -5/2$; therefore it is monotone on the interval $[-1/\sqrt{3}, 1/\sqrt{3}]$. So we can simply check the endpoints to verify that $y^2 + 5y + 5 > 1$ when $y \in [-1/\sqrt{3}, 1/\sqrt{3}]$, and hence it can never equal $\cos{x}$ if the first equation is satisfied.

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