Finding $f(\pi/3)$ where $f(x,y)=u(x,y)+iv(x,y)$

Question

Given that $f(x,y)=u(x,y)+iv(x,y)$ is an entire function of $z=x+iy$ such that $f(0)=-1$, $\partial u/\partial x=(e^y+e^{-y})\text{cos}x $ and $\partial u/\partial y=(e^y-e^{-y})\text{sin}x $, what is $f(\pi/3)$?

My Attempt: Since function is given entire, implies analytic everywhere. So, using C.R. conditions I found out $u(x,y)=(e^y+e^{-y})\text{sin}x$ and $v(x,y)=(e^y-e^{-y})\text{cos}x$. But couldn't proceed further as $f(x,y)$ has two variables but we are given $f(0)=-1$ and asked about $f(\pi/3)$ which are both in just one variable.

Please help me to solve this question. Thanks in advance.

Answer 1

After following JonathanZ suggestion I could solve the above question. Posting the complete solution here: $\partial u/\partial x=(e^y+e^{-y})\text{cos}x \implies u(x,y)=(e^y+e^{-y})\text{cos}x+g(y)$. Now since $\partial u/\partial y=(e^y-e^{-y})\text{sin}x $, differentiating $u(x,y)$ w.r.t $y$ to get $g(y)=c$.

So, $u(x,y)=(e^y+e^{-y})\text{cos}x+c$

Since $\partial u/\partial x=\partial v/\partial y$ and $\partial u/\partial y=-\partial v/\partial x$. On solving we get $v(x,y)=(e^y-e^{-y})\text{cos}x+k$.

Now using the initial condition $f(0)=-1$, we find that $c=-1\;,k=0$ and so, $$f(x,y)=[(e^y+e^{-y})\text{sin}x-1]+i[(e^y-e^{-y})\text{cos}x]$$ Now $f(\pi/3)=f(\pi/3,0)=2\times\frac{\sqrt3}{2}-1+i(0)=\sqrt{3}-1$.

$$f(\pi/3)=\sqrt3-1.$$

Answer 2

It's pretty clear that you haven't figured out the most general form of $f$, as you could add an arbitrary constant and $f$ would still satisfy the partial derivative requirements. But it can be more complicated than that. Recall that when you "integrate away" a regular derivative, you pick up an arbitrary constant:

$$df/dx = x \implies f(x) = x^2/2 + C $$

When you "integrate away" a partial derivative, you pick up an arbitrary function of the other variable

$$\partial f/\partial x = xy^2 \implies f(x,y) = x^2y^2/2 + g(y)$$

(You can also think of it as picking up an arbitrary constant, but the arbitrary constant can be different for different values of $y$, hence a function of $y$.)

You should also see that when you "integrate away" the $\partial/\partial y$ you pick up a function of $x$. Once you've done that correctly you can substitute back in to the C-R equations and solve for $g(y)$ and $h(x)$, at least up to an arbitrary constant, which $f(0)$ should give you.

I think the confusion between $f(x,y)$ and $f(\pi/3)$ is just the author being lazy about the distinction between $f(x,y)$ and $f(z)$, and $\pi/3$ is just the complex number $\pi/3 + 0\cdot i$.