Unlock the Secrets of Complex Analysis: Why Taylor Expansions Are a Breeze in the Complex Plane

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In the complex case, if a function is differentiable in the sense of complex variables then the validity of the Taylor expansion of that function is guaranteed. In this sense, complex analysis is “simple” relative to the real case.

### Exercises

1. Suppose that the complex series \(\sum_{n=1}^{\infty}a_{n}\) converges. Show that \(\lim_{n\to\infty}a_{n}=0\).
2. Determine whether the given infinite series converges or diverges: 1. \(\sum_{n=1}^{\infty}\left(\frac{1+2i}{\sqrt{6}}\right)^{n}\) 2. Determine whether the given infinite series converges or diverges: 1. \(\sum_{n=1}^{\infty}\left(\frac{1+2i}{\sqrt{6}}\right)^{n}\) 2. \(\sum_{n=1}^{\infty}\frac{i^{n}}{n}\) 3. Show that each of the following series converges for all \(z\): \(\mbox{a)}\,\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\) b) \(\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n}}{(2n)!}\) c) \(\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n+1}}{(2n+1)!}\) 4. Prove that if \(a_{n}\) for \(n=0,1,\dots\) are complex numbers such that \(\sum|a_{n}|\) converges, then \[\left|\sum_{n=0}^{\infty}a_{n}\right|\leq\sum_{n=0}^{\infty}|a_{n}|.\]
5. By differentiating an appropriate series, show that \[\sum_{k=1}^{\infty}kz^{k}=\frac{z}{(1-z)^{2}}\quad\mbox{for}\quad|z|<1.\] 6. Write down an expansion of the form \(\sum_{n=0}^{\infty}a_{n}z^{n}\) for 1. \(\frac{1}{3z+4}\) 2. \(\frac{1+iz}{1-iz}\) 7. Show that if \(\{a_{n}\}\) is a sequence of nonzero complex numbers such that \[\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_{n}|}=L,\] then \[\lim_{n\to\infty}|a_{n}|^{\frac{1}{n}}=L.\] 8. Find a series expansion of \((1-z)^{-1}\) valid in \(|z+3|<4\).

9. Show that for \(a\neq b\),

\[\frac{1}{(z-a)(z-b)}=\frac{1}{(a-b)}\left(\sum_{n=0}^{\infty}\left(\frac{1}{b^{n+1 }}-\frac{1}{a^{n+1}}\right)z^{n}\right)\]

is true for \(|z|<\min\{|a|,|b|\}\).

10. Prove the following:

1. The power series \(\sum\frac{z^{n}}{n^{2}}\) converges at every point of the unit circle.
2. The power series \(\sum nz^{n}\) does not converge on any point of the unit circle.
3. The power series \(\sum\frac{z^{n}}{n}\) converges at every point of the unit circle except at \(z=1\).

11. Determine all holomorphic functions \(f\) defined on the unit disc \(D\) which satisfy

\[f^{\prime\prime}\left(\frac{1}{n}\right)+f\left(\frac{1}{n}\right)=0.\]

12. Let the sequence \(a_{0},a_{1},\dots\) be defined such that

\[1-x^{2}+x^{4}-x^{6}+\dots=\sum_{n=0}^{\infty}a_{n}(x-3)^{n}\quad 0

Find \(\mbox{limsup}_{n\to\infty}\left(|a_{n}|^{\frac{1}{n}}\right)\).

13. Suppose \(f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\) has a radius of convergence \(R>0\). Show that \(g(z)=\sum_{n=0}^{\infty}\frac{a_{n}z^{n}}{n!}\) is a holomorphic function on \(\mathbb{C}\) and that for \(0

\[|g(z)|\leq Me^{\frac{|z|}{r}}.\]

### 3.4 Some Holomorphic Functions

#### The Exponential Function

In the section on power series we already showed that the power series \(\sum\frac{z^{n}}{n!}\) has an infinite radius of convergence and we may define the _exponential function_as:

\[e^{z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\ \ \ \ \mbox{ for }z\in\mathbb{C},\]

so \(e^{z}\) is holomorphic and hence also continuous in \(\mathbb{C}\). For \(z=x+iy\),

\[e^{z}=e^{x+iy}=e^{x}(\cos y+i\sin y),\]

and we see that

\[f(z)=e^{z}=u(x,y)+iv(x,y),\]

where

\[u(x,y)=e^{x}\cos y\mbox{ and }v(x,y)=e^{x}\sin y.\]

The functions \(u(x,y)\) and \(v(x,y)\) are continuous and real valued with continuous first-order partial derivatives and satisfy the Cauchy-Riemann equations,

\[u_{x}=e^{x}\cos y=v_{y}\mbox{ and }u_{y}=-e^{x}\sin y=-v_{x},\]

\[\frac{d}{dz}(e^{z})=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x }=e^{x}\cos y+ie^{x}\sin y=e^{z}.\]

**Modulus, Argument, and Conjugate of \(e^{z}\)**:

\[\mbox{If }\,w=e^{z}=e^{x}e^{iy}\mbox{ then }|w|=|e^{z}|=e^{x}\mbox{ and }\,\arg(e^{z})=y+2k\pi.\]

Since \(e^{x}>0\) for all \(x\), we know \(|e^{z}|>0\). In fact, \(e^{z}\neq 0\) for all \(z\). The value \(w=0\) is not in the range of \(e^{z}\), however as we know \(e^{\pi i}=-1\), \(e^{z}\) could be negative. For the conjugate, it is easy to see that

\[\overline{e^{z}}=e^{\overline{z}}.\]

The following are some algebraic properties of \(e^{z}\). Let \(z_{1}\) and \(z_{2}\) be complex numbers, then

* \(e^{0}=1\).
* \(e^{z_{1}}\cdot e^{z_{2}}=e^{z_{1}+z_{2}}\).
* \(\frac{e^{z_{1}}}{e^{z_{2}}}=e^{z_{1}-z_{2}}\).
* \((e^{z_{1}})^{n}=e^{nz_{1}}\ \ \ \ \ \ n=0,\pm 1,\pm 2,\ldots\).
* \(e^{z}\) is a periodic function with period \(2\pi i\).

To show e) above, consider \(e^{z+2\pi i}=e^{z}\cdot e^{2\pi i}=e^{z}\). Repeating this process, one can say \(e^{z+4\pi i}=e^{z}\) and thus \(e^{z+2n\pi i}=e^{z}\) for \(n=0,\pm 1,\pm 2,\ldots\) which in turn asserts that \(e^{z}\) is not a one-to-one function. In other words,

\[e^{z_{1}}=e^{z_{2}}\iff z_{1}=z_{2}+2n\pi i\ \ (n\in\mathbb{Z}).\]To find the inverse function of \(e^{z}\), first we divide the complex plane into horizontal strips by taking

\[-\infty

This infinite strip is called the _fundamental region_ (Figure 3.15) and if we suppose all values of \(e^{z}\) are assumed in the fundamental region then \(e^{z}\) is one-to-one.

Furthermore, if we look at a vertical line segment, say \(z=a+it\), \(-\pi

\[w=e^{z}=e^{a+it}=e^{a}\cdot e^{it}\]

is a circle centered at \(0\) radius \(e^{a}\) (Figure 3.16).

Figure 3.16: Image of a vertical line segment under \(w=e^{z}\)

Figure 3.15: Fundamental region

Images of the vertical line segments in \((-\pi,\pi]\) are circles.

On the other hand if we take a horizontal line in the fundamental region, the image of this line under \(w=e^{z}\) is a ray (Figure 3.17).

Since the only \(z\) on the horizontal line \(y=b\) is of the form \(z=x+ib\) where \(-\infty

\[w=e^{z}=e^{x+ib}=e^{x}\cdot e^{ib}.\]

We define a new parameter \(e^{x}=s\) and observe that \(0

\[w=se^{ib}\quad\text{where}\quad 0

which is the set consisting of all points \(w\neq 0\) in the ray starting from \(0\) containing the point \(e^{ib}=\cos b+i\sin b\).