Unlock the Power of Exponents: 4 Challenging Exercises to Test Your Complex Analysis Skills

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### Exercises

1. Find the real and imaginary parts of \(e^{e^{z}}\).
2. Simplify \(e^{z^{2}}\), \(e^{iz}\), \(e^{\frac{1}{z}},z\neq 0\) where \(z=x+iy\).

2. Suppose \(f\) is holomorphic in \(D(0\,;1)\). Prove that if \(e^{f}\) is constant, then \(f\) is constant.

3. Find the image of the given set under the exponential mapping:

1. The line \(x=3\).
2. The infinite strip \(1

4. Establish the following identities: 1. \(\sin(z+w)=\sin z\cos w+\cos z\sin w\) for \(z,w\in\mathbb{C}\). 2. \(\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y\). 3. \(\sin(-z)=-\sin z\), \(\cos(-z)=\cos z\), \(\sin\left(\frac{\pi}{2}-z\right)=\cos z\). 4. \(\sin\overline{z}=\overline{\sin z}\), \(e^{\overline{z}}=\overline{e^{\overline{z}}}\).
5. Use the equation \(\sin z=\sin x\cosh y+i\sinh y\cos x\), where \(z=x+iy\), to prove that \[|\sinh y|\leq|\sin z|\leq|\cosh y|.\]
6. Find all solutions to the following equations :

1. \(\cosh z=-1\).
2. \(\tan z=i\).
7. Find all values of: 1. \(\log 1\) and \(\log i\).
2. \(\log z\), where \(z=\dfrac{5+5\sqrt{3}i}{2}\).
8. Compute the following values when the \(\log\) is defined by its principal value on the open set \(S\) equal to the plane with the positive real axis deleted.

1. \(\log(-1+i)\).
2. \((-1)^{i}\).

9. Solve \(\sin z=5\).

10. Show that for \(z_{1},z_{2}\in\mathbb{C}\setminus\{0\}\), the identity for the principal value of the logarithm

\[\log(z_{1}z_{2})=\log z_{1}+\log z_{2}\]

holds true up to the integral multiples of \(2\pi i\).

11. Find a domain on which \(\log(3z-i)\) is holomorphic.

12.

12. Verify that \((z^{\alpha})^{n}=z^{n\alpha}\) for \(z\neq 0\) and \(n\) an integer.
2. Give an example to show that for \(z\neq 0\), we can have \((z^{\alpha_{1}})^{\alpha_{2}}\neq z^{\alpha_{1}\alpha_{2}}\).

### 3.5 Conformal Mappings

Suppose that \(w=f(z)\) is a complex mapping defined on an open set \(G\). We say the mapping \(f\) is _conformal_ at the point \(z_{0}\) if it “preserves the angle” between any two curves intersecting at \(z_{0}\). As we shall see later, every holomorphic function whose derivative is nonzero defines such mappings. These mappings, apart from their geometric interest, are used in advanced complex analysis, as well as solving certain physical problems such as problems in two-dimensional fluid flow.( See, e.g., [54], p. 263 and p. 386). The main motivation behind conformal mappings is to transform a given region \(G\) onto a geometrically simpler region \(G^{*}\), say the unit disc \(D\) or upper half-plane \(\mathbb{H}\), where many ideas have been developed to study holomorphic functions. In the following we will consider mapping a region whose boundary consists of lines, rays, circular arcs, and line segments. We will see that for many such cases, all we need is the following three basic types of maps and their certain linear combinations, namely:

1. Linear fractional transformations (also known as Mobius transformations)
2. Exponentials
3. Integer and non-integer powers

Before we understand angle preserving maps, we need to clarify what we mean by the angle between two curves \(\gamma_{1}\) and \(\gamma_{2}\).

### Angle Between Curves

Assume that \(\gamma_{1}\) and \(\gamma_{2}\) are smooth curves in \(G\) that intersect at \(z_{0}\) with the parametrization

\[\gamma_{1}(t)=(x_{1}(t),y_{1}(t))\quad\text{and}\quad\gamma_{2}(t)=(x_{2}(t),y _{2}(t)).\]

Suppose the orientation of both curves are in the direction of increasing \(t\), and since \(z_{0}=\gamma_{1}(t_{0})=\gamma_{2}(t_{0})\), the angle between these two curves is defined to be the angle between the tangent vectors to the curves (Figure 3.21).

Note that tangent vectors of the curves are

\[\frac{d\gamma_{1}}{dt}=\left(\frac{dx_{1}}{dt},\frac{dy_{1}}{dt}\right)\quad\text {and}\quad\frac{d\gamma_{2}}{dt}=\left(\frac{dx_{2}}{dt},\frac{dy_{2}}{dt} \right),\]

and we are interested in the dot product between the tangent vectors \(\frac{d\gamma_{1}}{dt}\) and \(\frac{d\gamma_{2}}{dt}\). Also the angle between \(\gamma_{1}\) and \(\gamma_{2}\) satisfies

\[\theta=\arg\gamma_{1}^{\prime}(z_{0})-\arg\gamma_{2}^{\prime}(z_{0})\]

in the interval \([0,\pi]\).

**Remark 69**.: In general, the cosine of the angle between \(z=\gamma_{1}\) and \(w=\gamma_{2}\) can be found by considering the dot product of these two vectors. For given two complex numbers \(z,w\) define the dot product by:

\[\langle z,w\rangle=\operatorname{Re}(z\overline{w}). \tag{3.1} \]\]

Then if \(\theta\) is the angle between \(z\) and \(w\),

\[\cos\theta=\frac{\langle z,w\rangle}{|z||w|}\quad\text{and}\quad\sin\theta= \frac{\langle z,-iw\rangle}{|z||w|}\]

determines the angle.

Figure 3.21: Angle between curves

**Definition 105**.: Let \(w=f(z)\) be a complex function defined in a domain \(G\) and let \(z_{0}\) be a point in \(G\). Then we say \(w=f(z)\) is _conformal at \(z_{0}\)_ if for every pair of smooth curves \(\gamma_{1}\) and \(\gamma_{2}\) intersecting at \(z_{0}\) the angle between \(\gamma_{1}\) and \(\gamma_{2}\) is equal to the angle between the image curves \(f(\gamma_{1})\) and \(f(\gamma_{2})\) at \(f(z_{0})\) (Figure 3.22).

**Theorem 126**.: _(Conformality Theorem) A holomorphic function \(w=f(z)\) whose derivative at \(z_{0}\) is nonzero will be conformal at \(z_{0}\)._

Proof

: Let the angle between \(\gamma_{1}\) and \(\gamma_{2}\) be \(\theta=\arg\gamma_{1}^{\prime}(z_{0})-\arg\gamma_{2}^{\prime}(z_{0})\). The curves \(\gamma_{1}\) and \(\gamma_{2}\) are mapped by \(f\) to paths \(f\circ\gamma_{1}\) and \(f\circ\gamma_{2}\); they meet at \(f(z_{0})\) at an angle

\[\phi=\arg(f\circ\gamma_{1})^{\prime}(z_{0})-\arg(f\circ\gamma_{2})^{\prime}(z _{0}).\]

By the chain rule

\[(f\circ\gamma_{1})^{\prime}(z_{0})=f^{\prime}(\gamma_{1}(z_{0}))\,\gamma_{1} ^{\prime}(z_{0})\quad\text{ and }\quad(f\circ\gamma_{2})^{\prime}(z_{0})=f^{\prime}(\gamma_{2}(z_{0}))\, \gamma_{2}^{\prime}(z_{0}).\]

Since both \(\gamma_{1}\) and \(\gamma_{2}\) are smooth, both \(\gamma_{1}^{\prime}\) and \(\gamma_{2}^{\prime}\) are nonzero. Furthermore, by the hypothesis \(f^{\prime}(z_{0})\neq 0\) and \(\gamma_{1}(z_{0})=\gamma_{2}(z_{0})\), the assertion of the theorem follows from

\[\frac{\gamma_{1}^{\prime}(z_{0})}{\gamma_{2}^{\prime}(z_{0})}=\frac{(f\circ \gamma_{1})^{\prime}(z_{0})}{(f\circ\gamma_{2})^{\prime}(z_{0})}\]

and the fact that \(\arg\left(\frac{z}{w}\right)=\arg(z)-\arg(w)\). Thus \(\phi=\theta\) in \([0,\pi]\).

**Remark 70**.: In the above proof, we could have used the scalar product given in (3.1) to prove the conformality theorem. Indeed if \(\alpha=f^{\prime}(z_{0})\neq 0\), then it can be shown that

\[\langle\alpha z,\alpha w\rangle=|\alpha|^{2}\langle z,w\rangle,\]

and thus the cosine of the angle between \(\alpha z\) and \(\alpha w\) is equal to the cosine of the angle between \(z\) and \(w\). The details are left to the reader.

Figure 3.22: Not conformal

**Example 101**.:
* The entire function \(f(z)=e^{z}\) is conformal at every point in the complex plane since \(f^{\prime}(z)=e^{z}\neq 0\) for all \(z\in\mathbb{C}\).
* The entire function \(g(z)=z^{2}\) is conformal when \(z\neq 0\) since \(g^{\prime}(z)=2z\neq 0\) when \(z\neq 0\). However it is not conformal at \(z=0\); if we consider \(f(z)=z^{2}=w\) it takes the rays \(\arg z=\theta\) and \(\arg z=\mu\) meeting at the angle \(\theta-\mu\) to rays \(\arg w=2\theta\) and \(\arg w=2\mu\) meeting at angle \(2(\theta-\mu)\) as shown in Figure 3.23.

**Remark 71** (Angle magnification at a critical point).: If a complex-valued function \(f\) is holomorphic at \(z_{0}\) and if \(f^{\prime}(z_{0})=0\) then we say \(z_{0}\) is a _critical point_ for \(f\). Although it does not follow from the above conformality theorem, it is true that holomorphic functions are not conformal at critical points due to some angle magnification at a critical point (see [54], p. 355). For example, the function \(f(z)=\sin z\) is entire (i.e., holomorphic on the entire complex plane) and its derivative \(f^{\prime}(z)=\cos z=0\) if and only if \(z=(2n+1)\frac{\pi}{2}\) for \(n=0,\pm 1,\pm 2,\dots\), so each of these point is a critical point of \(f\). Thus \(w=\sin z\) is a conformal mapping for all \(z\neq(2n+1)\frac{\pi}{2}\), \(n=0,\pm 1,\pm 2,\dots\).

### Conformal Equivalence: Riemann Mapping Theorem