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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?

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

1 Answer 1

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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 $x\in \mathbb{R}$, $y=\dfrac{\pi n}{2}$, $n\in \mathbb{Z}$.

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  • $\begingroup$ 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. $\endgroup$ Apr 29 at 14:35
  • $\begingroup$ Ah, I see. Well, too bad he was not communicating about it. $\endgroup$ Apr 29 at 16:24
  • $\begingroup$ Yeah. They should have commented on that. $\endgroup$ Apr 29 at 17:15

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