Complex Analysis

Representing Complex numbers with the Exponential function

We know that a complex number can be expressed in the following form, where z \in \C {C}:

 z= r(cos\theta + isin\theta)

Taking the derivative with respect to \theta shows that this representation of complex numbers leads to another, arguably more elegant representation:

 \frac{{d}}{{d} \theta} \left [ r(cos\theta + isin\theta) \right ]= r(-sin\theta +icos\theta)= ir(cos\theta+isin\theta)

Taking the derivative just multiplied z by a factor of i. It is then a trivial extension to say that taking further derivatives adds further multiplicative factors of i. Thus:

 \frac{{d^{n}}}{{d} \theta^{n}} \left [ r(cos\theta + isin\theta) \right ] = i^{n}r(cos\theta +isin\theta)

Now we know  \frac{{d^{n}}}{{d} \theta^{n}} for all n a Taylor expansion of z around 0 can be taken:

z = r \left [ z(0) + z'(0)\theta + z''(0)\frac{\theta^{2}}{2!}+ ... \right ]

The last thing that needs to be done is to evaluate the nth derivatives at 0, remembering that  sin(0) = 0, cos(0) = 1 :

z = r \left [ 1 + i\theta +\frac{i^{2}\theta^{2}}{2!}+ ...\right ] = r\sum \frac{(i\theta)^{n}}{n!}

This expansion looks very similar to another important Taylor expansion:

e^{x} = 1 + x + \frac{x^{2}}{2!}+ ... =\sum \frac{(x)^{n}}{n!}

This shows that the complex number z can be represented in the form re^{i\theta}.

Functions of Complex Variables

-Powers of Complex Numbers

De Moivre’s Theorem states that for z \in \C {C}:

 z^{n}= r^{n}e^{in\theta} = r^{n} \left [ cos n\theta + isin n\theta \right ]

Thus powers of complex numbers can be evaluated.

-Roots of Complex Numbers

Consider the solutions to  z^{\frac{1}{4}} = 81 , we can write 81 in exponential form so (argument is 2k\pi as each integer multiple, k,  of 2\pi is equivalent to the principal argument 2\pi):

 z^{4}= 81e^{i2k\pi}

Taking the 4th root of both side (using de Moivre’s theorem to evaluate a fractional power) gives:

 z= 81^{\frac{1}{4}}e^{\frac{ik\pi}{2}}

This means that the modulus of the roots (r') will be 81^{\frac{1}{4}}=3, and the argument will be \frac{ik\pi}{2}, 1\leqslant n\leqslant 4, as there will be 4 roots.

In general if we have the following:

 z^{n}= \rho e^{i \phi}


 z= \rho^{\frac{1}{n}} e^{i (\frac{\phi}{n}+\frac{2k\pi}{n})} = \rho^{\frac{1}{n}} \left [ cos (\frac{\phi}{n}+\frac{2k\pi}{n}) +isin(\frac{\phi}{n}+\frac{2k\pi}{n}) \right ]

-Complex Power Series

A function defined by the real variable,  x can be expressed in terms of a series in the form (see Taylor expansion):

 f(x) = f(0) + f'(0)x + \frac{f''(0)x^{2}}{2!}+ ... + \frac{f^{n}(0)x^{n}}{n!}

This means that if a function’s derivatives, and its value, are bounded at x=0 the series converges everywhere. This, however, is defined for real function, and real arguments, thus we need to extend it to the complex plane.

One of the ways to test whether an infinite series converges is the ratio test. The ratio used for this test, \rho, is defined in the following way:

    $$ \rho = \lim_{n\rightarrow \infty} \left | \frac{A_{n+1}}{A_{n}} \right | $$

The series will converge if \rho < 1, and it will diverge if \rho > 1. In the marginal case that \rho = 1, the test is inconclusive and some other method must be used. For example let us consider the following complex series, where z \in \C  {C}:

    $$ \sum_{n=0}^{\infty}z^{n}= 1+z+z^{2}+ ... $$

In this case the ratio will be calculated in the following way:

\rho= \left | \frac{z^{n+1}}{z^{n}} \right |= \left | z \right |

This tells us that the series converges for  \left | z \right | < 1, the locus for which lies within the circle centred on the origin with unit radius. This is called the “circle of convergence”  for the series.

A very important series in complex analysis is the Maclaurin series for e^{z}:

    $$e^{z}= \sum_{n=0}^{\infty} \frac{z^{n}}{n!}$$

We can do the ratio test for this series to see whether it converges:

    $$ \rho=\lim_{n\rightarrow \infty} \left | \frac{z^{n+1}n!}{z^{n}(n+1)!} \right |=\lim_{n\rightarrow \infty} \frac{\left | z \right |}{n+1}= 0 $$

This means that \rho < 1 for all complex numbers in the plane, meaning that the radius of convergence is infinite. Thus e^{z} can be formally defined by it’s Maclaurin series.

-Complex Exponential Function

The complex exponential function converges for all points in the complex plane, as shown above. It is now important to show that the complex exponential function has all the properties that the real exponential function- these can be shown by taking advantage of the series representation of the exponential function.

The first property to extend to the complex exponential function is that e^{z_{1}}e^{z_{2}}= e^{z_{1} + z_{2}}. We can demonstrate that this property extends to complex numbers by using the series representation for e^{z}:

    $$e^{z_{1}}e^{z_{2}}= \bigg ( \sum_{n=0}^{\infty} \frac{z_{1}^{n}}{n!} \bigg )\bigg ( \sum_{n=0}^{\infty} \frac{z_{2}^{n}}{n!} \bigg )$$

This requires the multiplication of two infinite series, and this is done by taking the “Cauchy product” of the two series:

    $$ \bigg ( \sum_{n=0}^{\infty} \frac{z_{1}^{n}}{n!} \bigg )\bigg ( \sum_{n=0}^{\infty} \frac{z_{2}^{n}}{n!} \bigg )=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{z_{1}^{k}}{k!}\frac{z_{2}^{n-k}}{(n-k)!}$$

Manipulating the inner sum yields an interesting result:

    $$ \frac{1}{n!} \sum_{k=0}^{n}\frac{n!}{k!(n-k)!} z_{1}^{k} z_{2}^{n-k}$$

We recognise that the inner sum is equal to one upon n! multiplied by the binomial expansion of z_{1}, z_{2}:

    $$(z_{1} + z_{2})^{n}=\frac{n!}{k!(n-k)!} z_{1}^{k} z_{2}^{n-k} $$

We can therefore conclude that:

    $$e^{z_{1}}e^{z_{2}}=\sum_{n=0}^{\infty} \frac{(z_{1} + z_{2})^{n}}{n!} = e^{z_{1} + z_{2}}$$


From this it is a trivial extension to show that (e^{z})^{n} = e^{nz},  thus we have extended the real exponential rules to complex numbers.

-Complex Trigonometric Function

We know the following, taking advantage of the even and odd nature of the trigonometric functions:

 e^{i\theta}= cos\theta + isin\theta

 e^{-i\theta}= cos\theta - isin\theta

From this summing and taking the difference between the two equations yields the following:

    $$ {cos\theta =\frac{e^{i\theta} + e^{-i\theta}}{2} $$

    $$ {sin\theta =\frac{e^{i\theta} - e^{-i\theta}}{2i} $$

Now consider the cosine function, but with a complex argument, z:

    $$ cos(z) = cos (x+iy) = \frac{e^{i(x+iy)}+ e^{-i(x+iy)}}{2} $$

    $$ = \frac{1}{2}(e^{ix}e^{-y} + e^{-ix}e^{y}) =\frac{1}{2} \left [ e^{-y}(cosx+isinx) + e^{y}(cosx-isinx) \right ] $$

With some simple manipulation the following is acheived:

    $$ cos(z) = cosx \bigg (\frac{e^{y} + e^{-y}}{2} \bigg ) - isinx \bigg (\frac{e^{y} - e^{-y}}{2} \bigg ) $$

The coefficients of the real and imaginary parts are familiar, they are the exponential definition for the hyperbolic cosine and sine functions respectively, thus:

    $$ cos(z) = cosxcoshy - isinxsinhy $$

This allows us to say the following about the real and imaginary parts of the cosine function with a complex argument:

$ Re(cos(z)) = cosxcoshy $

$ Im(cos(z)) = -sinxsinhy $

For the case where  x=0 and z is pure imaginary we get the following elegant result, showing that the hyperbolic cosine is analogous to the cosine of an imaginary angle! :


A similar technique can be used to determine the sine of a complex argument, this results in the following:

sin(z)= sinxcoshy + icosxsinhy

sin(iz)= isinhy

We can also use the results above to determine the magnitude of the complex cosine function:

 \left | cos(z) \right | = \sqrt{cos^{2}xcosh^{2}y + sin^{2}xsinh^{2}y}

 \left | cos(z) \right | = \sqrt{cos^{2}x(1+sinh^{2}y) + sin^{2}xsinh^{2}y}

 \left | cos(z) \right | = \sqrt{cos^{2}x + (cos^{2}x+sin^{2}x)sinh^{2}y} = \sqrt{cos^{2}x + sinh^{2}y}

Again, a similar method can be applied to find the magnitude of the complex sine function:

\left | sin(z) \right | =  \sqrt{sin^{2}x + sinh^{2}y}

Consider the magnitude of both the complex sine, and the complex cosine functions, as y \rightarrow \infty . As their magnitudes both depend on  sinhy they are thus both unbounded. If you consider the exponential form of  sinhy it is clear that as y \rightarrow \infty sinhy \rightarrow \infty , thus \left | cos(z) \right | \rightarrow \infty and \left | sin(z) \right | \rightarrow \infty , as \left | Im(z) \right | \rightarrow \infty .

-Complex Logarithm