Real and imaginary parts of a complex function [closed]

Is it always possible to separate the real and the imaginary parts of a complex function ? And why ? I always did it by calculations, but is there a theorem that says that the division in real and imaginary parts is always possible ?

closed as unclear what you're asking by Guy Fsone, Rohan, Martin R, Claude Leibovici, MoyaDec 12 '17 at 13:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• For the same reason that you can always get the $x$- and $y$-coordinate of a point. Complex numbers are "nothing more" than a different view of the 2D-plane with some fancy operations defined on them, like addition and multiplication. – M. Winter Dec 12 '17 at 9:58
• because those functions apply to complex give a complex number – Guy Fsone Dec 12 '17 at 10:01
• edited the main post – Poiera Dec 12 '17 at 10:01

Here is your theorem:

Theorem. Given a function $f:X\to \Bbb C$ from some arbitrary set $X$ into the complex number. Then there are functions $f_R,f_I:X\to\Bbb R$ so that $f=f_R+if_I$.

Proof.

Define

$$f_R=\frac 12[f+\bar f],\qquad f_I=\frac1{2i}[f-\bar f],$$

where $\bar f$ denotes the complex conjugate of $f$. We show that $f_R$ is real. Note that a complex number $z$ is real if and only if $\bar z=z$. Therefore

$$\overline{f_R}=\overline{1/2[f+\bar f]}=\frac12\overline{[f+\bar f]}=\frac12[\bar f+\bar{\bar f}]=\frac12[\bar f+f]=f_R,$$

sufficed to show that $f_R$ is real. The same works for $f_I$. It remains to show

$$f_R+if_I=\frac12[f+\bar f]+i\frac1{2i}[f-\bar f]=\frac12[f+\bar f+f-\bar f]=\frac12[2f]=f$$

and we are done. $\square$

Usually we denote

$$\mathrm{Re}(f)=f_R\qquad \text{and}\qquad \mathrm{Im}(f)=f_I.$$

Above theorem gives a handy way to compute these:

$$\mathrm{Re}(f)=\frac12[f+\bar f],\qquad \mathrm{Im}(f)=\frac1{2i}[f-\bar f],$$

assuming that you believe that the complex conjugate always exists.

Geometric intuition

Note that $\Bbb C$ is just another way to write $\Bbb R^2$, i.e. the 2D-plane of point $(x,y)$. Every complex number $z=x+iy$ can be associated with a point $(x,y)$. The real and imaginary part are the $x$- and $y$-coordinate respectively.

Think about a function $f:\Bbb R\to\Bbb C$. This is a complex function. But you can view it as a curve in the plane by associating $\Bbb C\cong\Bbb R^2$. This gives you $f:\Bbb R\to\Bbb R^2$, i.e. a curve. And it seems obvious that we can always extract $x$- and $y$-coordinate of a curve, right?

$$f(t)=(x(t),y(t))^\top.$$

The functions $\operatorname{Re}, \operatorname{Im}:\Bbb C\to \Bbb R$ are well-defined, and may, like any well-defined function that takes complex numbers as input, be composed with a complex function $f:\Bbb C\to \Bbb C$ to give the two functions $${\operatorname{Re}}\circ f:\Bbb C\to \Bbb R\\ {\operatorname{Im}}\circ f:\Bbb C\to \Bbb R$$ which extract the real part and the imaginary part of $f$, respectively.

• It's right using \operatorname, in general, but in this case it should be {\operatorname{Re}}\circ f in order to get the right spacing. – egreg Dec 12 '17 at 10:30
• @egreg Now that you say it, you're right. The naked \operatorname breaks the mathbin property of \circ. – Arthur Dec 12 '17 at 12:02

Yes.

Take $f:\mathbb C\to\mathbb C$.

Then, by definition, for every $z\in\mathbb C$, the value $f(z)$ is a complex number. Therefore, $f(z)$ can be written as $f(z)=a+bi$. The "real part" of $f$ simply maps $z$ to its corresponding $a$, and the "complex part" maps it to $b$.

More rigorously, you can define $$f_r(z) = \mathrm{Re}(f(z)) = \frac{f(z) + \overline{f(z)}}{2}$$ and

$$f_i(z)=\mathrm{Im}(z)=\frac{f(z)-\overline{f(z)}}{2i}$$

and you can show that $f_r$ and $f_i$ both take only real values and that $$f(z)=f_r(z)+if_i(z)$$

• Is it possible to connect this to the fact that, using Fourier series, we can "create" any function ? – Poiera Dec 12 '17 at 10:23
• @Poiera That's way beyonf the scope of the original question, and also way too vague a statement to carry any meaning. – 5xum Dec 12 '17 at 10:23

Yes, it can be done.

Remember that andy complex variable function can be written as $f(z) = f(x + iy)$, and this may provide, or mayn't, a simple (or less) possibility to decompose it into $g(x) + ih(y)$, or more generally and better:

$$f(z) = f(x + iy) = \Re(f(z)) + \Im(f(z))$$

Simple Examples

1)

$$e^z = e^{x + iy} = e^x + e^{iy} = \sinh(x) + \cosh(x) + \cos(y) + i\sin(y) = \Re(e^z) + \Im(e^z)$$

2)

$$\sin(z) = \sin(x + iy) = \sin(x)\cos(iy) + \sin(iy)\cos(x) = \sin(x)\cosh(y) + i\sinh(y)\cos(x)$$

That is,

$$\sin(x)\cosh(y) + i(\sinh(y)\cos(x)) = \Re(\sin(z)) + \Im(\sin(z))$$

Less Simple Examples

$$\sqrt{z} = \sqrt{x + iy} = z^{1/2}$$

Here you need the complex numbers root evaluation, or the exponentialization

$$z^{1/2} = \large e^{1/2\ln(z)}$$

Then you will deal with the logarithm

$$\log(z) = \log(|z|) + i\arg(z)$$

Then we always may find monstrous examples but it's not the case.

For a rigorous answer, check 5xsum's one!

take a function $f:\Bbb C\mapsto\Bbb C$.

this function is $f(z)=w$. because we know from the definition of $f$ that both $z,w$ are complex i can rewrite it: $f(a+ib)=c+id$, but $c,d$ are not constant(well they can, but it doesn't matter), they depends on $z$, if they depends on $z$ it means that they are functions, hence $f(z)=c(z)+id(z)$.

more formal proof will be $f(z)$ has imaginary and real part, $\overline {f(z)}$ has the same real part and the opposite imaginary part. so ${f(z)}+\overline {f(z)}=2\Re(f(z))\implies \Re(f(z))=\frac{{f(z)}+\overline {f(z)}}2$ and the same we can do ${f(z)}-\overline {f(z)}=2i\Im(f(z))\implies \Im(f(z))= \frac{f(z)-\overline{f(z)} }{2i}$

the notation of $\Im$ is the imaginary part and $\Re$ is the real part

you can look at it as $g:\Bbb R^2\mapsto\Bbb R^2, g(a,b)=(c,d)$. here again we can see that $c$ and $d$ are depends on the input of the function, so we can write it as $g(a,b)=(c(a,b),d(a,b))$.