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Chapter 32: Problem 11

Sind f眉r die Funktion \(f, \mathbb{C} \rightarrow \mathbb{C}\) mit $$ f(z)= \begin{cases}\frac{z^{5}}{\left.z z\right|^{4}} & \text { f眉r } z \neq 0 \\\ 0 & \text { f眉r } z=0\end{cases} $$ im Punkt \(z_{0}=0\) die Cauchy-Riemann-Gleichungen erf眉llt? Ist \(f\) in \(z_{0}=0\) komplex differenzierbar?

Short Answer

Expert verified
Based on the given complex function, we found that the Cauchy-Riemann equations are satisfied at \(z_0 = 0\). Therefore, the function is complex-differentiable at \(z_0 = 0\).

Step by step solution

01

Rewrite function in polar coordinates

Let \(z = re^{i\theta}\). We can write the function \(f\) in polar coordinates as $$ f(re^{i\theta})= \begin{cases} \frac{(re^{i\theta})^{5}}{r^{4}} & \text{for } r \neq 0 \\ 0 & \text{for } r=0\end{cases}. $$ Now we can simplify the expression for \(r \neq 0\): $$ f(re^{i\theta})= \begin{cases} r e^{5i\theta} & \text{for } r \neq 0 \\ 0 & \text{for } r=0\end{cases}. $$
02

Compute partial derivatives

Now we can find the partial derivatives of \(f\) with respect to \(x\) and \(y\) using the Cauchy-Riemann equations: $$ u(r,\theta) = \operatorname{Re}(f(re^{i\theta})) = r\cos(5\theta), $$ $$ v(r,\theta) = \operatorname{Im}(f(re^{i\theta})) = r\sin(5\theta). $$ We compute the partial derivatives: $$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x} = \cos(5\theta) - 5r\sin(5\theta) \frac{-y}{r^2}, $$ $$ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial y} = 5r\sin(5\theta) \frac{x}{r^2}, $$ $$ \frac{\partial v}{\partial x} = \frac{\partial v}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial v}{\partial \theta} \frac{\partial \theta}{\partial x} = 5r\cos(5\theta) \frac{x}{r^2}, $$ $$ \frac{\partial v}{\partial y} = \frac{\partial v}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial v}{\partial \theta} \frac{\partial \theta}{\partial y} = \sin(5\theta) - 5r\cos(5\theta) \frac{-x}{r^2}. $$
03

Check Cauchy-Riemann equations

Now we check whether the partial derivatives satisfy the Cauchy-Riemann equations at \(z_0 = 0\): $$ \frac{\partial u}{\partial x} (0,0) = 1, $$ $$ \frac{\partial v}{\partial y} (0,0) = 0, $$ $$ \frac{\partial u}{\partial y} (0,0) = 0, $$ $$ \frac{\partial v}{\partial x} (0,0) = 0. $$ Since \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\) at \(z_0 = 0\), the Cauchy-Riemann equations are satisfied.
04

Evaluate complex differentiability

As the Cauchy-Riemann equations are satisfied, the function \(f(z)\) is complex-differentiable at \(z_0 = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cauchy-Riemann Equations
The Cauchy-Riemann Equations are a crucial set of conditions in complex analysis. They determine whether a complex function is differentiable. These equations apply to a function expressed in terms of real-valued functions, say \(u(x,y)\) and \(v(x,y)\), representing the real and imaginary parts of a complex function \(f(z)\), where \(z = x + iy\). For \(f(z) = u(x,y) + iv(x,y)\) to be complex differentiable at a point, the following must be satisfied:
  • \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
  • \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)
These conditions ensure the function behaves well-differentiated in terms of its real and imaginary parts, akin to the concept of derivatives in real analysis. Meeting these criteria means the function is complex-differentiable at that point, a vital property for ensuring analytical functions behave predictably in calculus.
Complex Function
Complex functions extend the idea of functions into the complex plane. A complex function \(f: \mathbb{C} \rightarrow \mathbb{C}\) maps complex numbers to complex numbers, gaining richer behavior compared to real functions. In our exercise, the function \(f(z)\) sketched is piecewise defined.
For \(z eq 0\), its form is \( \frac{z^5}{|z|^4}\) and it is zero when \(z = 0\). Such functions can often require analysis through the geometry and algebra inherent in their complex nature.
Understanding how these functions behave involves breaking down these functions into their real part \(u(x,y)\) and imaginary part \(v(x,y)\), turning complex analysis tasks into more manageable steps.
Polar Coordinates
Polar coordinates offer a different perspective to analyze and simplify complex functions by expressing complex numbers in terms of their modulus and angle. For a point \((r, \theta)\), the complex number \(z\) is given as \(z = re^{i\theta}\), where \(r\) is the distance from the origin and \(\theta\) is the angle from the positive x-axis.
In the original exercise, converting to polar coordinates simplifies our function considerably. The transformation \(z= re^{i\theta}\) reduces \(\frac{z^5}{|z|^4}\) into \(re^{5i\theta}\), facilitating easier computation and manipulation. This capacity to convert into a trigonometric and exponential framework is invaluable for identifying symmetries and simplifying integration in complex spaces.
Partial Derivatives
Partial derivatives help dissect the changes in functions with respect to each of its variables independently. For real-valued functions, partial derivatives determine the rate of change in one direction while keeping other variables constant.
In the realm of complex functions, this becomes an essential tool when dealing with the Cauchy-Riemann equations. They require computing derivatives of real and imaginary parts separately with respect to \(x\) and \(y\). This step-by-step differentiation reveals if the components satisfy the requisite criteria for complex differentiability.
  • \( \frac{\partial u}{\partial x} \) and \( \frac{\partial v}{\partial y} \)
  • \( \frac{\partial u}{\partial y} \) and \(\frac{\partial v}{\partial x} \)
Evaluating these partial derivatives carefully ensures that the necessary conditions are met, verifying the differentiability of complex functions in the exercise.

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Most popular questions from this chapter

Man ermittle ohne Rechnung den Konvergenzradius bei Entwicklung der jeweils angegebenen Funktion um den Punkt \(z_{0}\) in eine Potenzreihe: (a) \(f(z)=\frac{z}{(z-\mathrm{i})(z+2)}\) um \(z_{0}=0\) (b) \(f(z)=\log (z) \operatorname{um} z_{0}=2+\mathrm{i}\) (c) \(f(z)=1 / \sin \left(\frac{1}{z}\right)\) um \(z_{0}=\frac{1}{\pi}+\mathrm{i}\)

Man zeige, dass die ,"H盲ufungspunktbedingung" im Identit盲tssatz tats盲chlich notwendig ist, dass also zwei holomorphe Funktionen, die auf einer unendlichen Menge \(M\) 眉bereinstimmen, nicht gleich sein m眉ssen, wenn \(M\) keinen H盲ufungspunkt hat.

Gibt es eine Funktion \(f(z)\) mit der Eigenschaft $$ f\left(\frac{1}{n}\right)=\frac{1}{1-\frac{1}{n}} \quad \text { f眉r } n=2,3,4, \ldots $$ die (a) auf \(|z|<1\), (b) auf ganz \(\mathbb{C}\) holomorph ist?

Man bestimme zu den folgenden Funktionen \(f\), \(D(f) \rightarrow \mathbb{C}\) jeweils die maximale Definitionsmenge und die Residuen der Funktionen an allen Singularit盲ten: (a) \(f(z)=\frac{\mathrm{e}^{n z}}{z^{2}+1}\) (b) \(f(z)=\frac{1}{z^{4}+2 z^{2}-3}\) (c) \(f(z)=\frac{4 z^{2}-5 z+3}{z^{3}-2 z^{2}+z}\)

Man berechne die Integrale $$ \begin{aligned} &I_{1}=\int_{0}^{\pi} \frac{\mathrm{e}^{i t}+1}{\mathrm{e}^{i t}+\mathrm{e}^{-i i}} \mathrm{~d} t \\ &I_{2}=\int_{0}^{1}\left(t^{3}+(\mathrm{i}+1) t^{2}+(\mathrm{i}-1) t+2 \mathrm{i}\right) \mathrm{d} t \\ &I_{3}=\int_{0}^{1} \frac{2 t}{t^{2}+(1+\mathrm{i}) t+\mathrm{i}} \mathrm{d} t \end{aligned} $$
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