# Determine at what points the complex function $f\left(z\right)=e^{2x}\cos3x+ie^{3x}\sin2y$ is differentiable.

Determine at what points the complex function $$f(z)=e^{2x}\cos3y+ie^{3x}\sin2y$$

is differentiable.

The function is differentiable at $$z_0 \in \mathbb C$$ if $$f$$ is defined on a neighborhood of $$z_0$$ contained in $$D_f=\mathbb C$$ (in this example this condition is satisfied) and $$\lim_{z \to z_0}\frac{f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}$$ exists, but here computing the limit is difficult, so what's the the alternative solution?

And generally when we are asked to find all points that a specific function is differentiable at such points what should we do?

• Do you know the Cauchy--Riemann equations? Do they hold here?
– Pedro
Apr 5 at 10:28
• @user1040538 According to your comment, there is a typo in your $f(z)$ that is $\cos 3y$. Apr 21 at 0:30
• I think the derivatives are still incorrect, the imaginary part is independent of x and the real part is independent of y. Apr 21 at 18:03
• I assume you're using $x$ and $y$ to denote the real and imaginary parts of $z$?
– Dan
Apr 21 at 18:20
• @user1040538 According to your comment in #2, do you mean $f\left(z\right)=e^{2x}\cos3y+ie^{3x}\sin2y$? Apr 22 at 1:04

We know $$f$$ is differentiable wherever the Cauchy-Riemann equations hold, that is, $$u_x=v_y$$ and $$u_y=-v_x$$
$$f\left(z\right)=e^{2x}\cos3x+ie^{3x}\sin2y$$, so let \begin{align*} u(x,y)&=e^{2x}\cos3x\\ v(x,y)&=e^{3x}\sin2y. \end{align*} Now we find \begin{align*} u_x&=\frac{d}{dx}(e^{2x}\cos3x)=2e^{2x}\cos 3x-3e^{2x} \sin 3x\\ v_y&=\frac{d}{dy}(e^{3x}\sin2y)=2e^{3x}\cos 2y\\ u_y&=\frac{d}{dy}(e^{2x}\cos3x)=0\\ v_x&=\frac{d}{dx}(e^{3x}\sin2y)=3e^{3x}\sin 2y, \end{align*} So we want to know where \begin{align*} 2e^{2x}\cos 3x-3e^{2x} \sin 3x&=2e^{3x}\cos 2y\\ 0&=-(3e^{3x}\sin 2y). \end{align*} For the first equation I used wolfram-alpha (because there is no point to waste your time doing this by hand) and got
$$x=\frac{1}{3} (2\pi n_1+\pi)$$, $$y=\frac{1}{2} (2 \pi n_2 - \cos^{-1}(-e^{(1/3(-2\pi n_1-\pi))}), n_1, n_2 \in \mathbb{Z}.$$
For the 2nd equation, we just need to find where $$\sin 2y=0$$, so we know that $$x\in \mathbb{R}$$, $$y=\dfrac{\pi n}{2}$$, $$n\in \mathbb{Z}$$.
• The version in the question body has $\cos 3y$ in $u$. I suspect a typo in the question title that went unnoticed. This may also explain why the asker was not happy with your solution. Apr 29 at 14:35