Quick question:

Quite often, when doing stuff in Complex Analysis, I'm asked to put something of the form $f = u+iv$ into the form $f(z)$. I HATE this step, because it always amounts to me just looking at it, and trying to sorta guess half way, and work backwards from the guess. It's time-consuming, tedious and clumsy. This will not do.

I'm now looking for appropriate theory (in the ideal case, an algorithm), or some other simple approach that will remove much of the guess work, remove stuff like the 'magically noticing' obscure trig identities, and just make the whole thing a bit more palatable.

What are some good tactics?

Thanks in advance.

  • 2
    $\begingroup$ Try replace $y$ with $0$ and $x$ with $z$ , works sometimes for me. $\endgroup$
    – Someone
    Jun 16, 2015 at 13:12
  • $\begingroup$ Closely related, almost a duplicate: math.stackexchange.com/questions/1173325 $\endgroup$
    – mrf
    Jun 16, 2015 at 14:03

2 Answers 2


Firstly I always check if it satisfies Cauchy-Riemann equations or not. If it satisfies you can write $f(x,y) = u(x,y)+iv(x,y)$ into the form $f(z)$. If it does not satisfy Cauchy-Riemann equations, you cannot write the $f(x,y) = u(x,y)+iv(x,y)$ into the form $f(z)$.

I would like to show my strategy in some examples

Example 1: ( you cannot convert the $f(x,y)$ into $f(z)$ in this example because it does not satisfy Cauchy-Riemann equations. Thus you do not need to struggle for converting into $f(z)$ because you cannot in any way)

$$f(x,y)=x^2+y^2+i2xy$$ $$u(x,y)=x^2+y^2$$ $$v(x,y)=2xy$$

Cauchy-Riemann equations:



$$2x=2x$$ $$2y \neq -2y$$ Thus you cannot do transform for example 1.

Example 2: (The example you can convert the $f(x,y)$ as $f(z)$)

$$f(x,y)=\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$$ $$u(x,y)=\frac{x}{x^2+y^2}$$ $$v(x,y)=-\frac{y}{x^2+y^2}$$

If you check, It satisfies Cauchy-Riemann equations thus you can convert into form of $f(z)$ Thus use the known relation for $z$ for convertion .

We know $z=x+iy$ so $x=z-iy$

Put it into the equation and you will see that $y$ will disappear after operations because it satisfies Cauchy-Riemann equations

$$f(z)=\frac{z-iy}{(z-iy)^2+y^2}-\frac{iy}{(z-iy)^2+y^2}$$ $$f(z)=\frac{z-iy}{z^2-2izy}-\frac{iy}{z^2-2izy}$$ $$f(z)=\frac{z-2iy}{z^2-2izy}=\frac{z-2iy}{z(z-2iy)}=\frac{1}{z}$$

  • $\begingroup$ Thank you for this! This will save me hours! :) $\endgroup$
    – JuliusL33t
    Jun 16, 2015 at 15:33
  • $\begingroup$ Putting $y=0$, $x=z$ and referring to the identity theorem is a lot easier. $\endgroup$
    – mrf
    Jun 17, 2015 at 7:06

Substitute $$ x=\frac{z+\bar z}{2},\quad y=\frac{z-\bar z}{2\,i}. $$ All $\bar z$ should desappear, leaving you with an expression deppending only on $z$.


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