2.1 Complex Functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle }$

We introduce functions of a complex variable.

1 Complex Functions

A complex function has the form $w = f(z)$ where $z$ and $w$ are complex variables. Let $u(x,y) = \Rep (f)$ and $v(x,y) = \Imp (f)$ . Then we can write $w = f(z) = u(x,y) + iv(x,y)$

Example 1 Let $w = f(z) = z^2$ . Find $u(x,y)$ and $v(x,y)$ .
Since $z = x+iy$ , $z^2 = (x+iy)^2 = \left (x^2 - y^2\right ) + i(2xy)$ Thus $u(x,y) = x^2 - y^2$ and $v(x,y) = 2xy$ .

(Problem 1a) Find the real and imaginary parts of the function $w = f(z) = \overline {z}$ $u(x,y) = \answer {x} \;\; \mbox {and} \;\; v(x,y) = \answer {-y}$

(Problem 1b) Find the real and imaginary parts of the function $w = f(z) = 2iz+i$ $u(x,y) = \answer {-2y} \;\; \mbox {and} \;\; v(x,y) = \answer {2x+1}$

(Problem 1c) Find the real and imaginary parts of the function $w = f(z) = z^3$ $u(x,y) = \answer {x^3 - 3xy^2} \;\; \mbox {and} \;\; v(x,y) = \answer {3x^2y -y^3}$

Here is a video solution to problem 1c:

(Problem 1d) Find the real and imaginary parts of the function $w = f(z) = 1/z$ $u(x,y) = \answer {\frac {x}{x^2 + y^2}} \;\; \mbox {and} \;\; v(x,y) = \answer {-\frac {y}{x^2 + y^2}}$

2 The Complex Exponential

To define $e^z$ we will use the power series representations of $e^x, \sin (x)$ and $\cos (x)$ : $e^x = \sum _{n=0}^\infty \frac {x^n}{n!},$ $\sin (x) = \sum _{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!},$ and $\cos (x) = \sum _{n=0}^\infty (-1)^n \frac {x^{2n}}{(2n)!}.$

We would like the complex exponential function to preserve as many properties of the real exponential function as possible. In particular, we would like the complex exponential to have the fundamental property $e^{z_1}\cdot e^{z_2} = e^{z_1 + z_2}$ . In particular, we would like $e^z = e^{x+iy} = e^x \cdot e^{iy}$ This reduces the challenge to defining $e^{iy}$ . To do this we make a “formal” power series expansion: \begin{align*} e^{iy} &= \sum _{n=0}^\infty \frac{(iy)^n}{n!} = \sum _{n=0}^\infty \frac{i^ny^n}{n!}\\ &= 1+iy -\frac{y^2}{2!} -i\frac{y^3}{3!} + \frac{y^4}{4!} + i\frac{y^5}{5!} - \frac{y^6}{6!} -i \frac{y^7}{7!} + \cdots \\ &= \sum _{n=0}^\infty (-1)^n\frac{y^{2n}}{(2n)!} + i\sum _{n=0}^\infty (-1)^n\frac{y^{2n+1}}{(2n+1)!}\\ &= \cos (y) + i\sin (y)\\ &= \cis (y) \end{align*}

In light of this calculation, we define the complex exponential function as $e^z \equiv e^x \cis (y)$ where $z = x+iy$ .

Example 2- Euler’s Identity Consider the function $f(z) = e^z$ . Compute $f(\pi i)$ .
Writing $\pi i$ as $0+\pi i$ , we use the definition of $e^z$ with $x = 0$ and $y= \pi$ . We have

$f(\pi i) = e^{\pi i} = e^0 \cis \pi = \cos (\pi ) + i\sin (\pi ) = -1$

This gives us Euler’s identity, $e^{\pi i} +1 = 0$

(Problem 2) Calculate each of the following:
i) $e^0 = \answer {1}$
ii) $e^{\pi i /4} = \answer {1/\sqrt 2 + i/\sqrt 2}$
iii) $e^{2\pi i}= \answer {1}$
iv) $e^3 = \answer {e^3}$
v) $e^{3+2\pi i} = \answer {e^3}$

Since $e^{2\pi i} = 1$ , the function $e^z$ is $2\pi i$ periodic.

Since $e^z = e^x \cis y$ , we have $e^z = u(x,y) + iv(x,y)$ with $u(x,y) = e^x\cos (y) \;\; \mbox {and} \;\; v(x,y)= e^x\sin (y)$

Example 3 Let $w = f(z) = e^{2z}$ . Find $u(x,y)$ and $v(x,y)$ .
Since $z = x+iy$ , $e^{2z} = e^{2x+2iy} = e^{2x}\cis (2y)$ Thus $u(x,y) = e^{2x}\cos (2y)$ and $v(x,y) = e^{2x}\sin (2y)$ .

(Problem 3a) Find the real and imaginary parts of the function $w = f(z) = e^{3z}$ $u(x,y) = \answer {e^{3x}\cos (3y)} \;\; \mbox {and} \;\; v(x,y) = \answer {e^{3x}\sin (3y)}$

(Problem 3b) Find the real and imaginary parts of the function $w = f(z) = e^{-z}$ $u(x,y) = \answer {e^{-x}\cos (y)} \;\; \mbox {and} \;\; v(x,y) = \answer {-e^{-x}\sin (y)}$

(Problem 3c) Find the real and imaginary parts of the function $w = f(z) = e^{\overline {z}}$ $u(x,y) = \answer {e^x\cos (y)} \;\; \mbox {and} \;\; v(x,y) = \answer {-e^x\sin (y)}$

Here is a video solution to problem 3c:

Example 4 Solve the equation $e^z = -4$ for $z$ .
Since $e^z = e^x \cis y$ and $-4 = 4\cis \pi$ , we have $e^x = 4 \;\; \mbox {and} \;\; y = \pi +2k\pi$ where $k \in Z$ . Thus $x = \ln 4 \;\; \mbox {and} \;\; y = (2k+1)\pi$

(Problem 4a) Solve for $z: e^z = -3$

There are infinitely many answers

Use $k$ to represent an arbitrary integer

$z = \answer {\ln 3 + i(2k\pi + \pi )}$

(Problem 4b) Solve for $z: e^z = i$

There are infinitely many answers

Use $k$ to represent an arbitrary integer

$z = \answer {i(2k\pi + \pi /2)}$

(Problem 4c) Solve for $z: e^z = 1+i$

There are infinitely many answers

Use $k$ to represent an arbitrary integer

$z = \answer {\ln \sqrt 2 +i(2k\pi + \pi /4)}$

Here is a video solution to problem 4c:

3 The Principal Argument Function

Recall the polar form of a complex number: $z = r\cis \theta$ where $r = |z|$ and $\theta$ is an angle co-terminal with the vector from $0$ to $z$ . Such an angle is called an argument of the complex number. If $\theta$ is an argument of $z$ , then any angle of the form $\theta +2k\pi$ where $k$ is an integer, is also an argument of $z$ .

We define the set $\arg (z) = \{\theta + 2k\pi | k \in \mathbb {Z}\}$ where $\theta$ is any argument of $z$ . To create an argument function, we must select one member of the set $\arg (z)$ for each $z \in \mathbb {C}$ . The principal argument function, $\Arg (z), z \neq 0$ , is defined by $\Arg (z) = \theta , \;\mbox {where}\; \theta \in \arg (z)\;\mbox {and}\; -\pi < \theta \leq \pi$

$\Arg (0)$ is not defined.

Example 5 Let $w = f(z) = \Arg (z)$ . Find $u(x,y)$ and $v(x,y)$ .
First, since $\Arg (z)$ is a real number, we have $u(x,y) = Arg(z)$ and $v(x,y) = 0$ . To find $u(x,y)$ we use the inverse tangent function. If $z=x+iy$ is in the first or fourth quadrant, then $\Arg (z) = \tan ^{-1}\left (\frac {y}{x}\right ).$ If $z$ is in the second quadrant, then $\Arg (z) = \tan ^{-1}\left (\frac {y}{x}\right ) + \pi .$ If $z$ is in the third quadrant, then $\Arg (z) = \tan ^{-1}\left (\frac {y}{x}\right ) - \pi .$ If $z =x$ is a real number, then $Arg(z) = 0$ if $x>0$ and $Arg(z) = \pi$ if $x<0$ .
If $z = iy$ is a purely imaginary number, then $Arg(z) = \pi /2$ if $y>0$ and $Arg(z) = -\pi /2$ if $y<0$ .

(Problem 5a) Find the following principal arguments:
$i) \Arg (\pi ) = \answer {0}$
$ii) \Arg (-2\pi ) = \answer {\pi }$
$iii) \Arg (\pi i) = \answer {\pi /2}$
$iv) \Arg (-i) = \answer {-\pi /2}$
$v) \Arg (1-i) = \answer {-\pi /4}$
$vi) \Arg (-1+i\sqrt 3) = \answer {2\pi /3}$

(Problem 5b) Recall that a geometric property of complex multiplication is the addition of angles: $z_1 \cdot z_2 = r_1 \cis \theta _1 \cdot r_2 \cis \theta _2 = r_1r_2\cis \left (\theta _1+\theta _2\right )$ Now, let’s consider $\Arg (z_1 \cdot z_2)$ and ruminate on the following question. Does the equation $\Arg (z_1\cdot z_2) = \Arg (z_1) + \Arg (z_2)$ hold for all non-zero complex numbers $z_1$ and $z_2$ ?

Yes No

Which of the following pairs of complex numbers can be used as a counter-example?

$1+i$ and $i-1$ $1+i$ and $1-i$ $1+i$ and $-1$ $-i$ and $-1-i$

4 Complex Roots

The Principal Argument function plays a role in creating a Principal $n^{th}$ root function. We begin with the square root. Just as there are two square roots of a positive real number, there are two square roots of a non-zero complex number. For the real square root, we distinguish these roots as positive and negative. However, this concept does not apply to non-real complex numbers. To find roots, we use the polar form $z = r\cis \theta = |z|e^{i\theta }$ .

Example 6 Find the two square roots of $4i$ .
The modulus of $4i$ is $4$ and the Principal Argument is $\pi /2$ , so in polar form $4i = 4\cis (\pi /2) = 4e^{\pi i/2}$ Since squaring a complex number squares the modulus and doubles the argument, to find a square root, we should take the (positive) square root of the modulus and halve the argument. Thus $\sqrt {4i} = 2\cis (\pi /4) = 2e^{\pi i/4} = \sqrt 2 + i\sqrt 2$ The other square root is the negative of this one, but we will compute it directly using a different argument for $4i$ . The idea of this method generalizes to $n^{th}$ roots. We can also write $4i$ in polar form as $4\cis (5\pi /2)$ . Thus the other square root is $\sqrt {4i} = 2\cis (5\pi /4) = -\sqrt 2 - i\sqrt 2$

Principal Square Root If $z \neq 0$ , the function $f(z) = z^{1/2} = |z|^{1/2} \cis \left (\frac {\Arg z}{2}\right ) = |z|^{1/2} e^{i\left (\frac {\Arg z}{2}\right )}$ is called the Principal Square Root function.

In the preceding example, since $\Arg 4i = \pi /2$ , the Principal Square Root is $(4i)^{\frac 12} = 2\cis (\pi /4) = 2e^{\pi i /4} = \sqrt 2 + i\sqrt 2$

(Problem 6) Find the Principal Square Root of the complex number. Write your answer in the form $a + bi$ .
i) $(2i)^{\frac 12} = \answer {1+i}$
ii) $(-144)^{\frac 12} = \answer {12i}$
iii) $(2 + 2i\sqrt 3)^{\frac 12} = \answer {\sqrt 3 + i}$
iv) $(-18+ 18i\sqrt 3)^{\frac 12} = \answer {3 + 3i\sqrt 3}$

The Principal $n^{th}$ Root function is defined by taking the positive $n^{th}$ root of the modulus and $(1/n)^{th}$ of the Principal Argument.

Principal $n^{th}$ Root function If $n$ is a positive integer and $z \neq 0$ , the function $f(z) = z^{\frac 1n} = |z|^{\frac 1n} \cis \left (\frac {\Arg z}{n}\right ) = |z|^{\frac 1n} e^{i\frac {\Arg (z)}{n}}$ is called the Principal $n^{th}$ Root function.

Example 7 Find the Principal Cube Root of $-8i$ and then find all of the other cube roots. Include them in a sketch.
Since $|-8i| = 8$ and $\Arg (-8i) = -\pi /2$ , we have $(-8i)^{\frac 13} = 8^{\frac 13} \cis (-\pi /6) = \sqrt 3 - i$ To find the other cube roots, we add multiples of $2\pi$ to the argument before dividing by $3$ . This gives \begin{align*} \sqrt [3]{-8i} &= 2 \cis \left [\frac 13\left (\frac{-\pi }{2} + 2k\pi \right )\right ] \\ &= 2 \cis \left (-\frac{\pi }{6} + \frac{2k\pi }{3}\right ) \end{align*}

This gives distinct values if $k = 0, 1$ or $2$ . If $k=0$ we get the Principal Cube Root. If $k = 1$ we get $\sqrt [3]{-8i}= 2 \cis \left (\frac {\pi }{2}\right ) = 2i$ and if $k = 2$ , we get $\sqrt [3]{-8i}= 2 \cis \left (\frac {7\pi }{6}\right ) = -\sqrt 3 - i$

(problem 7a) Find and sketch the 3 cube roots of each of the following:
\begin{align*} i) & \;1 \; \mbox{(these are called roots of unity)} \\ ii)& \;27 \\ iii) & \;-8 \\ iv) & \;i \end{align*}

(problem 7b) Find and sketch the 4 fourth roots of each of the following:
\begin{align*} i) &\; 1 \; \mbox{(these are called roots of unity)} \\ ii)&\; 16 \\ iii) &\; -81 \\ iv) & \;i \end{align*}

(problem 7c) Find and sketch the 5 fifth roots of each of the following:
\begin{align*} i) & \;1 \; \mbox{(these are called roots of unity)} \\ ii)& \;-32 \\ iii) & \;i \end{align*}

Here is a video solution to problem 7c, part iii:

2024-10-02 13:30:01