Unlock the Secrets of Complex Analysis: 3 Challenging Problems to Test Your Skills

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4. Show that \(f(z)=\overline{z}\) is not holomorphic at any point of \(\mathbb{C}\).

5. Show, by changing variables, that the Cauchy-Riemann equations in terms of polar coordinates become

\[\frac{\partial u}{\partial r}=\frac{1}{r}\,\frac{\partial v}{\partial\theta},\qquad\qquad\frac{\partial v}{\partial r}=-\frac{1}{r}\,\frac{\partial u}{ \partial\theta}.\]

6. Let \(f(z)\) be defined as:

\[f(z)=\frac{x^{4}-y^{4}}{x^{3}+y^{3}}+i\,\frac{x^{4}+y^{4}}{x^{3}+y^{3}}\qquad \text{ with }f(0)=0.\]

Show that \(f(z)\) is not differentiable at the origin, despite the fact that \(u\) and \(v\) satisfy the Cauchy-Riemann equations.

7. Suppose \(f(z)\) and \(\overline{f(z)}\) are holomorphic functions on an open set \(G\). Show that \(f(z)\) is constant in \(G\).

8. Show that if \(f(z)\) is holomorphic in an open set \(G\) and if one of the functions \(u(x,y)\), \(v(x,y)\) is constant, then the other one is also constant.

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If we use the convention that \(1/0=\infty\) and \(1/\infty=0\), then one can express \(R\) as _Hadamard’s formula_ (see [37], pp. 55-56):

\[\frac{1}{R}=\limsup|a_{n}|^{1/n}.\]

The region \(|z|

**Theorem 124**.: _Let \(\sum a_{n}z^{n}\) be a power series with radius of convergence \(0\leq R\leq\infty\). Then_

_(1)_ \(\sum a_{n}z^{n}\) _converges absolutely for all_ \(z\)_,_ \(|z|

_(2)_ \(\sum a_{n}z^{n}\) _diverges for any_ \(z\) _with_ \(|z|>R\)_._

_On the boundary of the disc of convergence, \(|z|=R\), a given power series either converges or diverges._

Proof

: To prove (1), if \(|z|0\), observing \(R-\epsilon\) is strictly less than \(R\) and using the definition of supremum, there is some \(z_{1}\) with

\[|z|=R-\epsilon<|z_{1}|\leq R\]

for which \(\sum|a_{n}z_{1}|\) converges. Then

\[|a_{n}z^{n}|\leq|a_{n}z_{1}^{n}|\]

for all \(n\) implies that \(\sum|a_{n}z^{n}|\) converges by the comparison test.

The proof of (2) is done by contradiction. Suppose there is some \(z\) with \(|z|>R\) for which \(\sum a_{n}z^{n}\) converges. Then the general term \(|a_{n}z^{n}|\) is bounded, i.e., \(\exists M\) such that \(|a_{n}z^{n}|\leq M\) for all \(n\). Now, pick \(z_{1}\) such that \(R<|z_{1}|<|z|\) so

\[|a_{n}z_{1}^{n}|=\left|a_{n}z^{n}\cdot\frac{z_{1}^{n}}{z^{n}}\right|\leq M \left|\frac{z_{1}}{z}\right|^{n}.\]

Since \(|z_{1}/z|<1\), the geometric series

\[\sum_{n=0}^{\infty}\left|\frac{z_{1}}{z}\right|^{n}\]

converges. Hence by the comparison test \(\sum|a_{n}z_{1}^{n}|\) converges. This contradicts the definition of \(R\).

In the following examples we see that \(R=1\) for the geometric series while \(R=\infty\) for the exponential function.

### Basic Examples

(1) The complex _exponential_ function, for \(z\in\mathbb{C}\), is defined as

\[e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}.\]

When \(z\) is real this definition coincides with the usual exponential function. Since

\[\left|\frac{z^{n}}{n!}\right|=\frac{|z|^{n}}{n!}\]

and

\[\sum_{n=0}^{\infty}\frac{|z|^{n}}{n!}=e^{|z|}<\infty,\]

the series above given for \(e^{z}\) converges absolutely for all \(z\in\mathbb{C}\).

It can be shown that \(e^{z}\) is holomorphic for all \(z\in\mathbb{C}\) (such functions are called _entire functions_) and its derivative is

\[\frac{d}{dz}(e^{z})=\sum_{n=0}^{\infty}\frac{nz^{n-1}}{n!}=\sum_{m=0}^{\infty }\frac{z^{m}}{m!}=e^{z}.\]

We are allowed to differentiate term by term since the series defining \(e^{z}\) is uniformly convergent in every disc in \(\mathbb{C}\).

(2) The trigonometric functions \(\sin z\) and \(\cos z\) can also be given by complex series that converge in the whole complex plane. These are

\[\sin z=\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n+1}}{(2n+1)!}\]

and

\[\cos z=\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n}}{(2n)!}.\]

Observe that the above expansions agree with the usual sine and cosine of the real argument. Using the fact that

\[e^{iz}=\cos z+i\sin z\]

and

\[e^{-iz}=\cos z-i\sin z,\]

one can obtain the so-called _Euler formulas_ for \(\sin z\) and \(\cos z\) as follows:

\[\sin z=\frac{e^{iz}-e^{-iz}}{2i}\,\ \cos z=\frac{e^{iz}+e^{-iz}}{2}.\](3) The _geometric series_\(\sum_{n=0}^{\infty}z^{n}\) converges absolutely only on the disc \(|z|<1\), and for those points in \(|z|<1\),

\[\sum_{n=0}^{\infty}z^{n}=\frac{1}{1-z},\]

and \(\frac{1}{1-z}\) is holomorphic in an open set \(D\setminus\{1\}\).