2

Might be a stupid question, but this has been bugging me. I'm revising my complex analysis course and I always come across writing $f(z)=u(x,y)+iv(x,y)$ to prove some theorem about complex functions, my question is: is this always possible? If yes, why? I can't think of a function for which this wouldn't be possible, but then I'm sure there are people who thought about it a lot more than me.

My idea is that this separation is always possible if the function is holomorphic at some point $z_0$, because then it would be possible to expand it as a Laurent series about that point, and since $(z-z_0)^k$ can be expanded in a sum of terms, one can then separate the real and imaginary part.

Does my reasoning hold? Is there a simpler way to think about it?

Thank you

user438666
  • 2,040
  • 1
  • 9
  • 23
  • 1
    The value depends on $z$ and $z$ depends on $x$ and $y$, so the value depends on $x$ and $y$. The value is a complex number and therefore has real and imaginary parts. – Mark Bennet May 24 '17 at 13:30

3 Answers3

1

HINT

Each complexed number could be uniquely denoted as $x+iy $ for $x,y \in \mathbb R$, and essentially $f(z)$ is a complexed number for each given $z \in \mathbb C$

Jay Zha
  • 7,792
  • That's obvious. I might have failed to properly formulate my question. When this came to my mind the first function that I thought of was $sin(z)$, but then I realized that I can write $sin(x+iy)=sin(x)cos(iy)+cos(x)sin(iy)$ and then rewrite the trigonometric functions of pure imaginaries as hyperbolic functions, my question is, how do I know that there isn't some crazy combination of functions for which this wouldn't be possible? – user438666 May 24 '17 at 13:29
  • @Chris so I've updated my answer a bit to answer your question. Essentially $f(z)$ is a complexed number given any input $z \in \mathbb C$, here we think real number is a special case of complex number when $y=0$. So you could alway express your function in such way. – Jay Zha May 24 '17 at 13:34
  • 1
    That was way more straightforward than I thought, thank you – user438666 May 24 '17 at 13:36
  • @Chris you are welcome! – Jay Zha May 24 '17 at 13:37
0

Every complex number can be decomposed uniquely into real and imaginary parts: $w = u + vi$ with $u$ and $v$ real. If $w = f(z)$ is an arbitrary complex-valued function, its real part $u$ and imaginary part $v$ satisfy $$ f(x, y) = u(x, y) + iv(x, y) $$ for all $(x, y)$.

Andrew D. Hwang
  • 78,195
0

If you have a function $f:\mathbb{C}\to\mathbb{C}$, then every element $z\in \mathbb{C}$ can be written uniquely (using the real and imaginary part) as $z=(x,y)=x+iy$ where $x=Re(z),y=Im(z)$ are real numbers.

This gives rise to a function $g:\mathbb{R}^2\to\mathbb{C}$ associated to $f$ such that $f(z)=f(x+iy)=g(x,y)$.

On the other hand, you can define $u(x,y)=Re(g(x,y))$ and $v(x,y)=Im(g(x,y))$, and we obtain the following: $$f(z)=f(x+iy)=g(x,y)=u(x,y)+i\cdot v(x,y)$$

Darío G
  • 4,878