Standard functions of a complex variable

1 Standard functions of a complex variable

The functions which we have considered so far have mostly been built from powers of $z$ . We consider other functions here.

1.1 The exponential function

Using Euler’s relation we are led to define

$e^{z} = e^{x + i y} = e^{x} . e^{i y} = e^{x} (cos y + i sin y) .$

From this definition we can show readily that when $y = 0$ then $e^{z}$ reduces to $e^{x}$ , as it should.

If, as usual, we express $w$ in real and imaginary parts then: $w = e^{z} = u + i v$ so that

$u = e^{x} cos y, v = e^{x} sin y$ . Then

$\frac{\partial u}{\partial x} = e^{x} cos y = \frac{\partial v}{\partial y} and \frac{\partial u}{\partial y} = - e^{x} sin y = - \frac{\partial v}{\partial x} .$

Thus by the Cauchy-Riemann equations, $e^{z}$ is analytic everywhere . It can be shown from the definition that if $f (z) = e^{z}$ then $f^{'} (z) = e^{z}$ , as expected.

Task!

By calculating $| e^{z} |^{2}$ show that $| e^{z} | = e^{x}$ .

$| e^{z} |^{2} = | e^{x} cos y + i e^{x} sin y |^{2} = {(e^{x} cos y)}^{2} + {(e^{x} sin y)}^{2} = {(e^{x})}^{2} ({cos}^{2} y + {sin}^{2} y) = {(e^{x})}^{2} .$

Therefore $|e^{z}| = e^{x}$ .

Solution

If $θ = arg (e^{z}) = arg (e^{x} (cos y + i sin y))$ then $tan θ = \frac{e^{x} sin y}{e^{x} cos y} = tan y$ . Hence $arg (e^{z}) = y$ .

Example 8

Find the solutions (for $z$ ) of the equation $e^{z} = i$

To find the solutions of the equation $e^{z} = i$ first write $i$ as $0 + 1 i$ so that, equating real and imaginary parts of $e^{z} = e^{x} (cos y + i sin y) = 0 + 1 i$ gives , $e^{x} cos y = 0$ and $e^{x} sin y = 1.$

Therefore $cos y = 0$ , which implies $y = \frac{π}{2} + k π$ , where $k$ is an integer. Then, using this we see that $sin y = \pm 1$ . But $e^{x}$ must be positive, so that $sin y = + 1$ and $e^{x} = 1$ . This last equation has just one solution, $x = 0$ . In order that $sin y = 1$ we deduce that $k$ must be even. Finally we have the complete solution to $e^{z} = i$ , namely:

$z = (\frac{π}{2} + k π) i$ , $k$ an even integer.

Task!

Obtain all the solutions to $e^{z} = - 1$ .

First find equations involving $e^{x} cos y$ and $e^{x} sin y$ :

As a first step to solving the equation $e^{z} = - 1$ obtain expressions for $e^{x} cos y$ and $e^{x} sin y$ from $e^{z} = e^{x} (cos y + i sin y) = - 1 + 0 i$ . Hence $e^{x} cos y = - 1$ , $e^{x} sin y = 0$ .

Now using the expression for $sin y$ deduce possible values for $y$ and hence from the first equation in $cos y$ select the values of $y$ satisfying both equations and deduce the form of the solutions for $z$ :

The two equations we have to solve are: $e^{x} cos y = - 1$ , $e^{x} sin y = 0$ . Since $e^{x} \neq 0$ we deduce $sin y = 0$ so that $y = k π$ , where $k$ is an integer. Then $cos y = \pm 1$ (depending as $k$ is even or odd). But $e^{x} \neq - 1$ so $e^{x} = 1$ leading to the only possible solution for $x$ : $x = 0$ . Then, from the second relation: $cos y = - 1$ so $k$ must be an odd integer. Finally, $z = k π i$ where $k$ is an odd integer. Note the interesting result that if $z = 0 + π i$ then $x = 0$ , $y = π$ and $e^{z} = 1 (cos π + i sin π) = - 1$ . Hence $e^{i π} = - 1$ , a remarkable equation relating fundamental numbers of mathematics in one relation.

1.2 Trigonometric functions

We denote the complex counterparts of the real trigonometric functions $cos x$ and $sin x$ by $cos z$ and $sin z$ and we define these functions by the relations:

$cos z \equiv \frac{1}{2} (e^{i z} + e^{- i z}), sin z \equiv \frac{1}{2 i} (e^{i z} - e^{- i z})$ .

These definitions are consistent with the definitions (Euler’s relations) used for $cos x$ and $sin x$ .

Other trigonometric functions can be defined in a way which parallels real variable functions. For example,

$tan z \equiv \frac{sin z}{cos z}$ .

Note that

$\frac{d}{d z} (sin z) = \frac{d}{d z} \{\frac{1}{2 i} (e^{i z} - e^{- i z})\} = \frac{1}{2 i} (i e^{i z} + i e^{- i z}) = \frac{1}{2} (e^{i z} + e^{- i z}) = cos z .$

Task!

Show that $\frac{d}{d z} (cos z) = - sin z .$

$\begin{array}{rcl} \frac{d}{d z} (cos z) & = & \frac{d}{d z} \{\frac{1}{2} (e^{i z} + e^{- i z})\} \\ = & \frac{i}{2} (e^{i z} - e^{- i z}) = - \frac{1}{2 i} (e^{i z} - e^{- i z}) = - sin z . \end{array}$

Among other useful relationships are

$\begin{array}{rcl} {sin}^{2} z + {cos}^{2} z & = & - \frac{1}{4} {(e^{i z} - e^{- i z})}^{2} + \frac{1}{4} {(e^{i z} + e^{- i z})}^{2} \\ = & \frac{1}{4} (- e^{2 i z} + 2 - e^{- 2 i z} + e^{2 i z} + 2 + e^{- 2 i z}) = \frac{1}{4} \cdot 4 = 1 . \end{array}$

Also, using standard trigonometric expansions:

$\begin{array}{rcl} sin z = sin (x + i y) = sin x cos i y + cos x sin i y & = & sin x (\frac{e^{- y} + e^{y}}{2}) + cos x (\frac{e^{- y} - e^{y}}{2 i}) \\ = & sin x cosh y - \frac{1}{i} cos x sinh y \\ = & sin x cosh y + i cos x sinh y . \end{array}$

Task!

Show that $cos z = cos x cosh y - i sin x sinh y .$

$\begin{array}{rcl} cos z = cos (x + i y) = cos x cos i y - sin x sin i y & = & cos x (\frac{e^{- y} + e^{y}}{2}) - sin x (\frac{e^{- y} - e^{y}}{2 i}) \\ = & cos x cosh y + \frac{1}{i} sin x sinh y \\ = & cos x cosh y - i sin x sinh y \end{array}$

1.3 Hyperbolic functions

In an obvious extension from their real variable counterparts we define functions $cosh z$ and $sinh z$ by the relations:

$cosh z = \frac{1}{2} (e^{z} + e^{- z}), sinh z = \frac{1}{2} (e^{z} - e^{- z})$ .

Note that $\frac{d}{d z} (sinh z) = \frac{1}{2} \frac{d}{d z} (e^{z} - e^{- z}) = \frac{1}{2} (e^{z} + e^{- z}) = cosh z$ .

Task!

Determine $\frac{d}{d z} (cosh z)$ .

$\frac{d}{d z} (cosh z) = \frac{1}{2} \frac{d}{d z} (e^{z} + e^{- z}) = \frac{1}{2} (e^{z} - e^{- z}) = sinh z .$

Other relationships parallel those for trigonometric functions. For example it can be shown that

$cosh z = cosh x cos y + i sinh x sin y$ and $sinh z = sinh x cos y + i cosh x sin y$

These relationships can be deduced from the general relations between trigonometric and hyperbolic functions (can you prove these?):

$cosh i z = cos z and sinh i z = i sin z$

Example 9

Show that ${cosh}^{2} z - {sinh}^{2} z = 1.$

Solution

$\begin{array}{rcl} {cosh}^{2} z & = & \frac{1}{4} {(e^{z} + e^{- z})}^{2} = \frac{1}{4} (e^{2 z} + 2 + e^{- 2 z}) \\ {sinh}^{2} z & = & \frac{1}{4} {(e^{z} - e^{- z})}^{2} = \frac{1}{4} (e^{2 z} - 2 + e^{- 2 z}) \\ ∴ {cosh}^{2} z & - & {sinh}^{2} z = \frac{1}{4} (2 + 2) = 1. \end{array}$

Alternatively since $cosh i z = cos z$ then $cosh z = cos i z$ and since $sinh i z = i sin z$ it follows that $sinh z = - i sin i z$ so that

${cosh}^{2} z - {sinh}^{2} z = {cos}^{2} i z + {sin}^{2} i z = 1$

1.4 Logarithmic function

Since the exponential function is one-to-one it possesses an inverse function, which we call $ln z$ . If $w = u + i v$ is a complex number such that $e^{w} = z$ then the logarithm function is defined through the statement: $w = ln z$ . To see what this means it will be convenient to express the complex number $z$ in exponential form as discussed in HELM booklet 10.3: $z = r e^{i θ}$ and so

$w = u + i v = ln (r e^{i θ}) = ln r + i θ .$

Therefore $u = ln r = ln |z|$ and $v = θ$ . However $e^{i (θ + 2 k π)} = e^{i θ} . e^{2 k π i} = e^{i θ} .1 = e^{i θ}$ for integer $k$ . This means that we must be more general and say that $v = θ + 2 k π$ , $k$ integer. If we take $k = 0$ and confine $v$ to the interval $- π < v \leq π$ , the corresponding value of $w$ is called the principal value of $ln z$ and is written Ln $(z)$ .

In general, to each value of $z \neq 0$ there are an infinite number of values of $ln z$ , each with the same real part. These values are partitioned into branches of range $2 π$ by considering in turn $k = 0$ , $k = \pm 1$ , $k = \pm 2$ etc. Each branch is defined on the whole $z -$ plane with the exception of the point $z = 0$ . On each branch the function $ln z$ is analytic with derivative $\frac{1}{z}$ except along the negative real axis (and at the origin). Figure 6 represents the situation schematically.

Figure 6

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The familiar properties of a logarithm apply to $ln z$ , except that in the case of Ln $(z)$ we have to adjust the argument by a multiple of $2 π$ to comply with $- π < arg (Ln (z)) \leq π$

For example

$ln (1 + i) = ln (\sqrt{2} e^{i \frac{π}{4}}) = ln \sqrt{2} + i (\frac{π}{4} + 2 k π)$
$= \frac{1}{2} ln 2 + i (\frac{π}{4} + 2 k π) .$
Ln $(1 + i) = \frac{1}{2} ln 2 + i \frac{π}{4}$ .
If $ln z = 1 - i π$ then $z = e^{1 - i π} = e^{1} . e^{- i π} = - e$ .

Task!

Find

$ln (1 - i)$
Ln $(1 - i)$
$z$ when $ln z = 1 + i π$

$ln (1 - i) = ln (\sqrt{2} e^{- i \frac{π}{4}}) = ln \sqrt{2} + i (- \frac{π}{4} + 2 k π) = \frac{1}{2} ln 2 + (- \frac{π}{4} + 2 k π)$ .
Ln $(1 - i) = \frac{1}{2} ln 2 - i \frac{π}{4}$ .
$z = e^{1 + i π} = e^{1} . e^{i π} = - e$ .

Exercises

Obtain all the solutions to $e^{z} = 1$ .
Show that $1 + {tan}^{2} z \equiv {sec}^{2} z$
Show that ${cosh}^{2} z + {sinh}^{2} z \equiv cosh 2 z$
Find $ln (\sqrt{3} + i), Ln (\sqrt{3} + i)$ .
Find $z$ when $ln z = 2 + π i$

$e^{x} cos y = 1$ and $e^{x} sin y = 0$ $∴ sin y = 0$ and $y = k π$ where $k$ is an integer.
Then $cos y = \pm 1$ and since $e^{x} > 0$ we take $cos y = 1$ and $e^{x} = 1$ so that $x = 0$ . Then $cos y = 1$ and $k$ is an even integer. $∴ z = 2 k π i$ for $k$ integer.
$tan z = \frac{1}{i} (\frac{e^{i z} - e^{- i z}}{e^{i z} + e^{- i z}})$ $\begin{array}{rcl} 1 + {tan}^{2} z = 1 - \frac{e^{2 i z} + e^{- 2 i z} - 2}{e^{2 i z} + e^{- 2 i z} + 2} = \frac{4}{e^{2 i z} + e^{- 2 i z} + 2} & = & \frac{2^{2}}{{(e^{i z} + e^{- i z})}^{2}} = \frac{1}{{cos}^{2} z} = {sec}^{2} z . \end{array}$
${cosh}^{2} z + {sinh}^{2} z = \frac{1}{4} (e^{2 z} + 2 + e^{- 2 z}) + \frac{1}{4} (e^{2 z} - 2 + e^{- 2 z}) = \frac{1}{2} (e^{2 z} + e^{- 2 z}) = cosh 2 z .$
$ln (\sqrt{3} + 1) = ln \sqrt{5} + i (\frac{π}{6} + 2 k π) = \frac{1}{2} ln 5 + i (\frac{π}{6} + 2 k π)$ . $Ln (\sqrt{3} + i) = \frac{1}{2} ln 5 + i \frac{π}{6}$ .
If $ln z = 2 + π i$ then $z = e^{2 + π i} = e^{2} e^{i π} = - e^{2}$ .

1 Standard functions of a complex variable

1.1 The exponential function

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1.2 Trigonometric functions

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1.3 Hyperbolic functions

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1.4 Logarithmic function

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