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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 ?

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

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    $\begingroup$ 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. $\endgroup$ – M. Winter Dec 12 '17 at 9:58
  • $\begingroup$ because those functions apply to complex give a complex number $\endgroup$ – Guy Fsone Dec 12 '17 at 10:01
  • $\begingroup$ edited the main post $\endgroup$ – Poiera Dec 12 '17 at 10:01
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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?

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

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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.

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  • $\begingroup$ 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. $\endgroup$ – egreg Dec 12 '17 at 10:30
  • $\begingroup$ @egreg Now that you say it, you're right. The naked \operatorname breaks the mathbin property of \circ. $\endgroup$ – Arthur Dec 12 '17 at 12:02
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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 $$f(z)=f_r(z)+if_i(z)$$

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

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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 $g(a,b)=(c(a,b),d(a,b))$.

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