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

${\text{e}}^z={\text{e}}^{x+\text{i}y}={\text{e}}^x.{\text{e}}^{\text{i}y}={\text{e}}^x\left(cosy+\text{i}siny\right).$

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

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

$u={\text{e}}^{x}cosy,v={\text{e}}^{x}siny$ . Then

$\frac{\partial u}{\partial x}={\text{e}}^{x}cosy=\frac{\partial v}{\partial y}\text{and}\frac{\partial u}{\partial y}=-{\text{e}}^{x}siny=-\frac{\partial v}{\partial x}.$

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

Task!

By calculating $|{\text{e}}^z{|}^2$ show that $\left|{\text{e}}^z\right|={\text{e}}^x$ .

Answer

Find $arg\left({\text{e}}^z\right)$ .

Solution

If $\theta =arg\left({\text{e}}^z\right)=arg\left({\text{e}}^x\left(cosy+\text{i}siny\right)\right)$  then  $tan\theta =\frac{{\text{e}}^{x}siny}{{\text{e}}^{x}cosy}=tany$ . Hence $arg\left({\text{e}}^z\right)=y$ .

Example 8

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

Solution

To find the solutions of the equation ${\text{e}}^z=\text{i}$ first write $\text{i}$ as $0+1\text{i}$ so that, equating real and imaginary parts of  ${\text{e}}^z={\text{e}}^x\left(cosy+\text{i}siny\right)=0+1\text{i}$ gives , ${\text{e}}^{x}cosy=0$ and ${\text{e}}^{x}siny=1.$

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

$z=\left(\frac{\pi }{2}+k\pi \right)\text{i}$ , $k$ an even integer.

Task!

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

First find equations involving ${\text{e}}^{x}cosy$ and ${\text{e}}^{x}siny$ :

Answer

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

Answer

We denote the complex counterparts of the real trigonometric functions $cosx$ and $sinx$ by $cosz$ and $sinz$ and we define these functions by the relations:

$cosz\equiv \frac{1}{2}\left({\text{e}}^{\text{i}z}+{\text{e}}^{-\text{i}z}\right),sinz\equiv \frac{1}{2\text{i}}\left({\text{e}}^{\text{i}z}-{\text{e}}^{-\text{i}z}\right)$ .

These definitions are consistent with the definitions (Euler’s relations) used for $cosx$ and $sinx$ .

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

$tanz\equiv \frac{sinz}{cosz}$ .

Note that

$\frac{d}{dz}\left(sinz\right)=\frac{d}{dz}\left\{\frac{1}{2\text{i}}\left({\text{e}}^{\text{i}z}-{\text{e}}^{-\text{i}z}\right)\right\}=\frac{1}{2\text{i}}\left(\text{i}{\text{e}}^{\text{i}z}+\text{i}{\text{e}}^{-\text{i}z}\right)=\frac{1}{2}\left({\text{e}}^{\text{i}z}+{\text{e}}^{-\text{i}z}\right)=cosz.$

Task!

Show that $\frac{d}{dz}\left(cosz\right)=-sinz.$

Answer

Also, using standard trigonometric expansions:

Task!

Show that $cosz=cosxcoshy-\text{i}sinxsinhy.$

Answer

1.3 Hyperbolic functions

In an obvious extension from their real variable counterparts we define functions $coshz$ and $sinhz$ by the relations:

$coshz=\frac{1}{2}\left({\text{e}}^z+{\text{e}}^{-z}\right),sinhz=\frac{1}{2}\left({\text{e}}^z-{\text{e}}^{-z}\right)$ .

Note that $\frac{d}{dz}\left(sinhz\right)=\frac{1}{2}\frac{d}{dz}\left({\text{e}}^z-{\text{e}}^{-z}\right)=\frac{1}{2}\left({\text{e}}^z+{\text{e}}^{-z}\right)=coshz$ .

Task!

Determine $\frac{d}{dz}\left(coshz\right)$ .

Answer

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

$coshz=coshxcosy+\text{i}sinhxsiny$ and $sinhz=sinhxcosy+\text{i}coshxsiny$

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

$cosh\text{i}z=cosz\text{and}sinh\text{i}z=\text{i}sinz$

Example 9

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

Solution

Alternatively since $cosh\text{i}z=cosz$ then $coshz=cos\text{i}z$ and since $sinh\text{i}z=\text{i}sinz$ it follows that $sinhz=-\text{i}sin\text{i}z$ so that

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

1.4 Logarithmic function

Since the exponential function is one-to-one it possesses an inverse function, which we call $lnz$ . If $w=u+\text{i}v$ is a complex number such that ${\text{e}}^w=z$ then the logarithm function is defined through the statement: $w=lnz$ . 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: $z=r{\text{e}}^{\text{i}\theta }$ and so

$w=u+\text{i}v=ln\left(r{\text{e}}^{\text{i}\theta }\right)=lnr+\text{i}\theta .$

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

In general, to each value of $z\ne 0$ there are an infinite number of values of $lnz$ , each with the same real part. These values are partitioned into branches of range $2\pi$ 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 $lnz$ 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

The familiar properties of a logarithm apply to $lnz$ , except that in the case of Ln $\left(z\right)$ we have to adjust the argument by a multiple of $2\pi$ to comply with $-\pi <arg\left(\text{Ln}\left(z\right)\right)\le \pi$

For example

1. $ln\left(1+\text{i}\right)=ln\left(\sqrt{2}{\text{e}}^{\text{i}\frac{\pi }{4}}\right)=ln\sqrt{2}+\text{i}\left(\frac{\pi }{4}+2k\pi \right)$

   $=\frac{1}{2}ln2+\text{i}\left(\frac{\pi }{4}+2k\pi \right).$

2. Ln $\left(1+\text{i}\right)=\frac{1}{2}ln2+\text{i}\frac{\pi }{4}$ .
3. If $lnz=1-\text{i}\pi$ then $z={\text{e}}^{1-\text{i}\pi }={\text{e}}^1.{\text{e}}^{-\text{i}\pi }=-e$ .

Task!

Find

1. $ln\left(1-\text{i}\right)$
2. Ln $\left(1-\text{i}\right)$
3. $z$ when $lnz=1+\text{i}\pi$

Answer

Exercises

1. Obtain all the solutions to ${\text{e}}^z=1$ .
2. Show that $1+{tan}^{2}z\equiv {sec}^{2}z$
3. Show that ${cosh}^{2}z+{sinh}^{2}z\equiv cosh2z$
4. Find $ln\left(\sqrt{3}+\text{i}\right),\text{Ln}\left(\sqrt{3}+\text{i}\right)$ .
5. Find $z$ when $lnz=2+\pi \text{i}$

Answer