Definition 1.6.1. Complex function.
A complex function is a function \(f\colon D\rightarrow \C\text{,}\) where \(D\subseteq \C\text{.}\)
Section 1.6 Complex functions
We now introduce the main object of study in this course, complex functions, and begin developing what ends up being an extension of calculus to this setting, starting with the notion of a limit of a complex function.
Subsection Complex functions
Definition 1.6.2. Real and imaginary parts.
Given a complex function \(f\colon D\rightarrow \C\) with \(D\subseteq \C\text{,}\) we can write
\begin{equation*}
f(z)=u(z)+iv(z)\text{,}
\end{equation*}
where \(u(z)=\Re f(z)\) and \(v(z)=\Im f(z)\text{.}\) We call the functions \(u\) and \(v\) the the real and imaginary parts of \(f\text{,}\) respectively, and write \(f=u+iv\text{.}\)
Identifying complex numbers with pairs in \(\R^2\) in the usual manner, we can think of
\begin{align*}
u \amp =u(x+iy)=u(x,y) \amp v=v(x+iy)=v(x,y)
\end{align*}
as real-valued functions in the real variables \(x\) and \(y\text{.}\)
Convention 1.6.3. Real and imaginary parts.
Given a complex function \(f\colon D\rightarrow \C\text{,}\) if we write \(f=u+iv\) without further comment, it is assumed that \(u\) and \(v\) are the real and imaginary parts of \(f\text{.}\)
Remark 1.6.4. Real and imaginary parts.
Given a complex function \(f\colon D\rightarrow \C\text{,}\) \(f=u+iv\) its real and imaginary parts are real-valued functions with domain \(D\text{.}\)
Example 1.6.5. Real and imaginary parts.
Let \(f\colon \C\rightarrow \C\) be defined as \(f(z)=z^2\text{.}\)
- Find the real and imaginary parts of \(f\text{.}\)
- Decide whether \(f\) is injective.
- Compute \(\im f=f(\C)\text{.}\)
Solution.
- Letting \(z=x+iy\text{,}\) we have\begin{align*} f(z) \amp =z^2\\ \amp = (x^2-y^2)+i(2xy)\text{,} \end{align*}and thus\begin{align*} u(x,y) \amp = x^2-y^2 \amp v(x,y)=2xy\text{.} \end{align*}
- Since \(f(-z)=f(z)=z^2\) for all \(z\text{,}\) we see that \(f\) is not injective: as a concrete counterexample, \(f(-1)=f(1)=1\text{.}\)
- According to Theorem 1.3.8, every complex number \(w\) has a square-root. Thus for all \(w\in \C\) there is a \(z\in \C\) such that \(f(z)=w\text{.}\) This proves \(f\) is surjective.
Definition 1.6.6. Complex exponential function.
We define the (complex) exponential function \(\exp\colon \C\rightarrow \C\) as
\begin{equation*}
\exp(x+iy)=e^x(\cos y+i\sin y)\text{.}
\end{equation*}
We will also us the notation \(e^z\) to denote \(\exp(z)\text{.}\)
Remark 1.6.7. Euler’s formula.
At long last we have a convenient shorthand notation for complex elements of the form \(z=\cos\theta+i\sin\theta\in S^1\text{:}\) namely, we see that
\begin{equation*}
e^{i\theta}=e^{0+i\theta}=e^0(\cos\theta+i\sin\theta)=\cos\theta+i\sin\theta\text{.}
\end{equation*}
This equality is called Euler’s formula. A particularly famous consequence of Euler’s formula is given by Euler’s equation:
\begin{equation*}
e^{i \pi}+1=0\text{.}
\end{equation*}
Theorem 1.6.8. Exponential function properties.
- We have \(e^{z+w}=e^ze^w\) for all \(z,w\in \C\text{.}\)
- We have \(e^z=e^w\) if and only if \(\Re z=\Re w\) and \(\Im w=\Im z+2\pi k\) for some \(k\in \Z\text{.}\) In particular, \(\exp\) is not injective.
- \(\im \exp=\C-\{0\}\text{.}\)
Proof.
- Given complex numbers \(z\) and \(w\text{,}\) we have\begin{align*} e^{z+w} \amp= e^{\Re (z+w)}e^{i\Im(z+w)}\\ \amp = e^{\Re z+\Re w}e^{i(\Im z+\Im w)}\\ \amp = (e^{\Re z}e^{i \Im z})\cdot (e^{\Re w}e^{i\Im w}) \amp (\knowl{./knowl/xref/th_polar_mult.html}{\text{1.2.16}})\\ \amp = e^ze^w\text{.} \end{align*}
- We have\begin{align*} e^z = e^w \amp \iff e^{\Re z}e^{i\Im z}=e^{\Re w}e^{i\Im w}\\ \amp \iff e^{\Re z}=e^{\Re w} \text{ and } \Im w=\Im z+2\pi k \text{ for some } k\in \Z \amp (\knowl{./knowl/xref/th_polar_form.html}{\text{1.2.13}})\\ \amp \iff \Re z=\Re w \text{ and } \Im w=\Im z+2\pi k \text{ for some } k\in \Z \text{,} \end{align*}where the last equivalence follows from the fact that the real exponential function is injective.
- Given any \(w\ne 0\) we can write \(w\) in polar form as \(w=se^{i\theta}\) for some \(s=\abs{w}> 0\) and \(\theta\in \R\text{.}\) Letting\begin{equation*} z=\ln s+i\theta=\ln \abs{w}+i\theta\text{,} \end{equation*}we have\begin{align*} e^z \amp =e^{\ln s}e^{i\theta}\\ \amp = se^{i\theta} \amp (\text{real exp., log. props.}) \\ \amp = w\text{.} \end{align*}This shows that for all \(w\in \C-\{0\}\) there is a \(z\in \C\) such that \(\exp(z)=w\text{.}\) Thus \(\im \exp\supseteq \C-\{0\}\text{.}\) It is clear that \(0\notin \im \exp\text{,}\) and hence that \(\im \exp=\C-\{0\}\text{.}\)
Remark 1.6.9. Solving \(e^z=w\).
From our analysis in the proof of Theorem 1.6.8, we saw that given a nonzero \(w\in \C\text{,}\) the complex number
\begin{equation*}
z=\ln\abs{w}+i\Arg w\text{,}
\end{equation*}
satisfies \(e^z=w\text{.}\) It follows from (2) of Theorem 1.6.8 that the set of all \(z\) satisfying \(e^z=w\) is thus
\begin{equation*}
\{\ln\abs{w}+i(\Arg w+2\pi k\colon k\in \Z\}=\{\ln\abs{w}+i(\theta)\colon \theta\in \arg z\}\text{.}
\end{equation*}
As such, as a sort of shorthand, we can express this set as
\begin{equation*}
\ln\abs{w}+i\arg(w)\text{,}
\end{equation*}
where as always \(\arg(w)=\{\Arg w+2\pi k\colon k\in \Z\}\) is the set of all arguments of \(w\text{.}\) Note that the various algebraic operations in that expression must be appropriately interpreted as an operations on sets.
Subsection Visualizing complex functions
How do we visualize a complex function \(f\colon \C\rightarrow \C\text{?}\) Identifying \(\C\) with \(\R^2\text{,}\) we can think of \(f\) as a function \(f\colon \R^2\rightarrow \R^2\text{.}\) As with any function, the graph \(\Gamma_f\) of \(f\) is defined as the set of all input-output pairs \((z,f(z))\text{.}\) Identifying \(z=x+iy\) with the pair \((x,y)\) and \(f(z)=u(x,y)+iv(x,y)\) with \((u(x,y),v(x,y))\) we have
\begin{equation*}
\Gamma_f=\{(z,f(z))\in \C^2\colon z\in \C\}=\{(x,y,u(x,y),v(x,y))\in \R^4\colon (x,y)\in \R^2\}\text{.}
\end{equation*}
This is not such an easy object to visualize, being a subset of \(\R^4\text{.}\) Instead, thinking of \(f\) as a transformation of the complex plane, we can visualize how \(f\) transforms various useful subsets of \(\C\text{.}\) This is typically done by starting with two copies of the complex plane (often called the \(xy\)- and \(uv\)-planes), sketching curves or regions \(D\) in the \(xy\)-plane, and sketching their image \(f(D)\) in the \(uv\)-plane.
Example 1.6.10. Squaring transformation.
Define \(f\colon \C\rightarrow \C\) as \(f(z)=z^2\text{.}\) Sketch the given curves or regions in the complex \(xy\)-plane, as well as the image of these sets under \(f\) in the complex \(uv\)-plane.
- \(C_r=\{z\colon \abs{z}=r\}\text{,}\) \(r\geq 0\text{.}\)
- \(\displaystyle \ell_\theta=\{z\in \C\colon \Arg z=\theta\}\)
- \(R=\{z\colon r\leq \abs{z}\leq s, \Arg z+2\pi k\in [\alpha, \beta]\text{ for some } k\in \Z\}\text{,}\) \(0\leq r\leq s\text{,}\) \(\alpha\leq \beta\text{.}\)
- \(v_r=\{z\in \C\colon \Re z=r\}\text{,}\) \(r\in \R\)
- \(h_s=\{z\in \C\colon \Im z=s\}\text{,}\) \(s\in \R\text{.}\)
- \(R=\{z\in \C\colon p\leq \Re z\leq q, r\leq \Im z\leq s\}\text{,}\) \(p\leq q\text{,}\) \(r\leq s\)
Solution.
We have
\begin{align*}
f(z)=f(x+iy) \amp =x^2-y^2+2xy\, i\\
\amp = r^2(\cos 2\theta+i\sin 2\theta)\text{,}
\end{align*}
where \(r=\abs{z}\) and \(\theta=\Arg z\text{.}\) Both descriptions will come in handy below.
- Looking at the polar description of \(f\text{,}\) we see that \(f\) takes a point \(z\) on \(C_r\text{,}\) rotates it by \(\Arg z\) and scales it by \(r\text{.}\) It is clear then that \(f(C_r)=(C_{r_2})\text{.}\)
- From the polar description of \(f\text{,}\) we see that\begin{equation*} \arg z^2=\{2\Arg z+2\pi k \colon k\in \Z\}\text{,} \end{equation*}from whence it follows that \(\Arg z^2=2\Arg z=2\pi k\) for some \(k\in \Z\text{.}\) It is easy to see that \(\ell_\theta=\ell_{\theta'}\) if and only if \(\theta=\theta'+2\pi k\) for some \(k\in \Z\text{.}\) Thus \(f\) maps \(\ell_\theta\) to \(\ell_{2\theta}\text{.}\)
- From the two previous parts, we conclude that \(f\) maps \(R\) to\begin{equation*} f(R)=\{z\colon r^2\leq \abs{z}\leq s^2, \Arg z+2\pi k \in [2\alpha, 2\beta] \text{ for some } k\in\Z \}\text{.} \end{equation*}See Figure 1.6.12.
-
Given \(z=r+yi\in v_r\text{,}\) we have\begin{equation*} f(z)=(r^2-y^2)+2ry i, \text{.} \end{equation*}When \(r=0\text{,}\) we see easily that this \(f(z)\) lies in \((-\infty, 0]\text{,}\) and that \(f(v_r)=(-\infty, 0]\text{.}\)Assume \(r\ne 0\text{.}\) Setting \(u=r^2-y^2\) and \(v=2ry\text{,}\) we see (after a little algebra) that\begin{equation*} u=-\frac{1}{4r^2}v^2+r^2\text{.} \end{equation*}This equation defines a horizontal parabola extending to the left in the \(uv\)-plane, with vertex \((r^2, 0)\text{,}\) and it is easy to see that the image of \(v_r\) under \(f\) is this entire parabola.
- A similar analysis as above shows that \(f(h_s)=[0,\infty)\) if \(s=0\text{,}\) and the rightward opening horizontal parabola defined by\begin{equation*} u=\frac{1}{4s^2}v^2-s^2\text{.} \end{equation*}
- Lastly, from the previous two parts it now follows that \(f(R)\) is the region bounded by the parabolas \(f(v_p)\text{,}\) \(f(v_q)\text{,}\) \(f(h_r)\) and \(f(h_s)\text{.}\) See Figure 1.6.11.
Our analysis in Example 1.6.10 gives us a complete understanding of how \(f(z)=z^2\) transforms two different coordinate systems of the plane: the cartesian coordinate system (represented by the standard rectangular grid), and the polar coordinate system (represented by the polar grid).
Figure 1.6.11 summarizes the action of \(f\) on the standard rectangular grid: it maps the \(x\)- and \(y\)-axes to the negative and positive \(x\)-axes, respectively, and it maps all other horizontal and vertical lines to horizontal parabolas.
The action of \(f(z)=z^2\) on the polar grid is somewhat simpler to describe: it maps the circles \(C_r\) of radius \(r\) centered at the origin to the circle \(C_{r^2}\) obtained by squaring the radius; and it maps the ray \(\ell_{\theta}\) that makes an oriented angle of \(\theta\) with the \(x\)-axis, to the ray \(\ell_{2\theta}\) obtained by doubling this angle.
Example 1.6.13. Exponential transformation.
Define \(f\colon \C\rightarrow \C\) as \(f(z)=\exp z=e^z\text{.}\) Sketch the given curves or regions in the complex \(xy\)-plane, as well as the image of these sets under \(f\) in the complex \(uv\)-plane.
- \(v_r=\{z\in \C\colon \Re z=r\}\text{,}\) \(r\in \R\)
- \(h_s=\{z\in \C\colon \Im z=s\}\text{,}\) \(s\in \R\text{.}\)
- \(R=\{z\in \C\colon p\leq \Re z\leq q, r\leq \Im z\leq s\}\text{,}\) \(p\leq q\text{,}\) \(r\leq s\)
Solution.
Writing
\begin{equation*}
f(z)=f(x+iy)=e^xe^{iy}=e^{\Re z}e^{i\Im y}.
\end{equation*}
we see that
\begin{align*}
f(v_r) \amp = \{f(r+iy\colon y\in \R\}=\{e^re^{iy}\colon y\in \R\}\\
f(h_s) \amp = \{f(x+is\colon x\in \R\}=\{e^xe^{is}\colon x\in \R\}\text{.}
\end{align*}
From the first description we see that for fixed \(r\) the set \(f(v_r)\) is the set of all complex numbers with modulus \(e^r\text{:}\) i.e., \(f(v_r)=C_{e^r}\text{,}\) the circle of radius \(e^r\) centered at the origin.
From the second description we see that for fixed \(s\) the set \(f(h_s)\) is the set of all positive scalar multiples of \(e^{is}\text{.}\) (Positive since \(e^x\) is positive for all \(x\in \R\text{.}\)) This is precisely the ray \(\ell_s\) of all complex numbers \(z\) satisfying \(\Arg z=s+2\pi k\) for some \(k\in \Z\text{.}\)
It follows that \(f\) maps a rectangular region of the form
\begin{equation*}
R=\{z\in \C\colon p\leq \Re z\leq q, r\leq \Im z\leq s\}
\end{equation*}
to the polar rectangular region
\begin{equation*}
f(R)=\{z\in \C\colon e^p\leq \abs{z}\leq e^q, r\leq \Arg z+2\pi k\leq s \text{ for some } k\in \Z\}\text{.}
\end{equation*}
See Figure 1.6.14.
Our analysis in Example 1.6.13 can be summarized as follows: the exponential function maps the vertical lines \(x=r\) to the circles \(C_{e^r}\) of radius \(r\) centered at the origin; and it maps the horizontal lines \(y=s\) to the rays \(\ell_s\) starting at the origin that make an angle of \(s\) radians with the positive \(x\)-axis. It follows that a rectangular region of the form
\begin{equation*}
R=\{z\in \C\colon a\leq \Re z\leq b, c\leq \Im z\leq d\}
\end{equation*}
gets mapped to the polar rectangular region
\begin{equation*}
f(R)=\{z\in \C\colon e^{a}\leq \abs{z} \leq e^b, c\leq \Arg z+2\pi k\leq d \text{ for some } k\in \Z\}\text{.}
\end{equation*}
Lastly from Theorem 1.6.8 and our anaylis above, it follows that if we restrict \(f(z)=e^z\) to the infinite horizontal strip \(R=\{z\in \C\colon -\pi< \Im z < \pi\}\text{,}\) then \(f\) defines a bijective function
\begin{equation*}
f\colon R\rightarrow \C-(-\infty, 0]\text{.}
\end{equation*}
