How Do You Decompose a Complex Function into Real and Imaginary Parts?

[fleazo]

One last simple question about complex analysis...

Hi, sorry again for having made so many threads. I have one remaining question about complex analysis that I keep get confused on.


Say that I have some complex function h(z). Sometimes I am really confused how to break that down into h(z) = u(x,y) + iv(x,y)


So for example say I want to take the integral ∫z²dz from 0 --> 1+i.


How do I know the u(x,y) and v(x,y) real functions that h(z) it is made up of? I mean... what if I want to see if the function is holomorphic? Or exact? Or harmonic? Or anything like that. I need the u and v components. For some functions, I just know it by a formula (like e^z = e^xcos(y)+ie^xsin(y) but I am really at a loss when I am just handed any arbitrary function. how do I find it? In fact, I'm not even sure how the formula for the complex exponential function was found. Was it just by converting to polar coordinates? I'm very confused on this point

 

[wnvl]

Simply replace z by x+iy in h(z).

 

[fleazo]

I guess what I mean is I need to find the actual functions of real variables that make up h(z).


For example e^z like I mentioned. I can show that e^z=u(x,y) + iv(x,y) where u(x,y)=e^xcos(y) and v(x,y) = e^xsin(y). So then I can show e^z is analytic, etc.


I'm just confused sometimes how to find thsi representation. I know about parameterizing but sometimes I don't entirely understand how it works. Like say I want to take ∫z²dz across the straight line from 0 to 1+i. Well I know I could paraterize that "curve" to be z(t)=t+it and then I can solve the integral. But sometimes I have a hard time understanding why the parameterization doesn't effect the integral. I mean... if I just had z² by itself and the question was "is z² holomorphic?" Then could I use that same parameterization and solve the Cauchy Reimann equations and get the answer? Is that valid?

 

[wnvl]

I guess what I mean is I need to find the actual functions of real variables that make up h(z).


For example e^z like I mentioned. I can show that e^z=u(x,y) + iv(x,y) where u(x,y)=e^xcos(y) and v(x,y) = e^xsin(y). So then I can show e^z is analytic, etc.


I'm just confused sometimes how to find thsi representation. I know about parameterizing but sometimes I don't entirely understand how it works. Like say I want to take ∫z²dz across the straight line from 0 to 1+i. Well I know I could paraterize that "curve" to be z(t)=t+it and then I can solve the integral. But sometimes I have a hard time understanding why the parameterization doesn't effect the integral. I mean... if I just had z² by itself and the question was "is z² holomorphic?" Then could I use that same parameterization and solve the Cauchy Reimann equations and get the answer? Is that valid?

Don't know whether it is allways possible for a holomorphic function to get an analytic expression for u and v? In case of z² it is evident.

If you use a specific parametrisation to solve the Cauchy Reimann equations, then you do not prove that a function is holomorphic on the entire domain.

All polynomial functions in z with complex coefficients are holomorphic on C, and so are sine, cosine and the exponential function. (The trigonometric functions are in fact closely related to and can be defined via the exponential function using Euler's formula). The principal branch of the complex logarithm function is holomorphic on the set C \ {z ∈ R : z ≤ 0}. The square root function can be defined as e^1/2.log(z) and is therefore holomorphic wherever the logarithm log(z) is. The function 1/z is holomorphic on {z : z ≠ 0}.
As a consequence of the Cauchy–Riemann equations, a real-valued holomorphic function must be constant. Therefore, the absolute value of z, the argument of z, the real part of z and the imaginary part of z are not holomorphic. Another typical example of a continuous function which is not holomorphic is complex conjugation.

 

[Citan Uzuki]

Well, if you really want to find the real and imaginary parts of z², just write z=x+iy, and multiply out: (x+iy)² = x² + 2xiy + y² = (x² + y²) + i(2xy), so u(x, y) = x² + y² and v(x, y) = 2xy

That said, you're getting way too caught up in the Cauchy-Riemann equations. For most functions, you want to work with z directly, rather than splitting it up into the real and imaginary parts. So the proof that z^2 is holomorphic becomes very simple:

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}z}z^2 & = \lim_{w \rightarrow 0} \frac{(z+w)^2 - z^2}{w} \\ & = \lim_{w \rightarrow 0} \frac{z^2 + 2zw + w^2 - z^2}{w} \\ & = \lim_{w \rightarrow 0} \frac{2zw + w^2}{w} \\ & = \lim_{w \rightarrow 0} 2z + w \\ & = 2z \end{align*}$$

Which, you will note, is exactly the same proof that x^2 is differentiable in the real case. All of the rational functions can be shown to be holomorphic in exactly the same way.

I'm not even sure how the formula for the complex exponential function was found. Was it just by converting to polar coordinates? I'm very confused on this point

Well of course you know e^x+iy = e^x e^iy, so the only difficulty is in showing that eiy = cos y + i sin y. And the way you do that is you just put z=iy into the Taylor series for e^z and separate out the real and imaginary parts, and note that the real part is the Taylor series for cos y and the imaginary part is the Taylor series for sin y. A derivation can be found on the Wikipedia here: http://en.wikipedia.org/wiki/Euler's_formula#Using_power_series

But sometimes I have a hard time understanding why the parameterization doesn't effect the integral.

First of all, it's affect. Now, the simplest reason the parameterization doesn't affect the integral is because z² admits a complex antiderivative (namely z³/3), and the fundamental theorem of calculus holds. More generally, it can be shown that the parameterization of an integral doesn't affect the integral for any homomorphic function on a simply connected domain. This is basically a consequence of Cauchy's integral theorem -- if the parameterization did matter, then integrating along one path and then backwards across the other would yield a nonzero integral along a closed curve.

 

[fleazo]

hi, i just wanted to thank you all for the responses and taking the time to write them. it has helped me very much