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Chapter 9: Problem 14

Bestimmen Sie jeweils alle \(z \in \mathbb{C}\), die L枚sungen der folgenden Gleichung sind. (a) \(\cos \bar{z}=\overline{\cos z}\), (b) \(\mathrm{e}^{\mathrm{i} \bar{z}}=\overline{\mathrm{e}^{\mathrm{i} z}}\)

Short Answer

Expert verified
Question: Determine all complex numbers 'z' that satisfy the given equations, (a) \(\cos \bar{z}=\overline{\cos z}\) and (b) \(\mathrm{e}^{\mathrm{i} \bar{z}}=\overline{\mathrm{e}^{\mathrm{i} z}}\). Answer: For (a), the given equation is satisfied for all \(z = x + iy \in \mathbb{C}\). For (b), the given equation is satisfied for all complex numbers \(z = x + iy\) where \(x\) and \(y\) are real numbers, and \(y = n\pi\) for \(n \in \mathbb{Z}\).

Step by step solution

01

Find the complex conjugate of \(\cos z\)

Write the formula for the cosine of a complex number in terms of its real and imaginary parts: \(\cos(z) = \cos(x + iy) = \cos(x)\cosh(y) - i\cdot\sin(x)\sinh(y)\). Then, find the complex conjugate of \(\cos(z)\): \(\overline{\cos(z)}=\cos(x)\cosh(y) + i\cdot\sin(x)\sinh(y)\).
02

Write the formula for \(\cos \bar{z}\)

The formula for the cosine of the complex conjugate of z is \(\cos(\bar{z}) = \cos(x - iy) = \cos(x)\cosh(y) + i\cdot\sin(x)\sinh(y)\).
03

Equate the expressions and solve for 'z'

Equate the expressions from Step 1 and Step 2: \(\cos(\bar{z}) = \overline{\cos(z)}\), which gives \(\cos(x)\cosh(y) + i\cdot\sin(x)\sinh(y)=\cos(x)\cosh(y) + i\cdot\sin(x)\sinh(y)\). Since these two expressions are equal, this equation is satisfied for all \(z = x + iy \in \mathbb{C}\). For (b) \(\mathrm{e}^{\mathrm{i} \bar{z}}=\overline{\mathrm{e}^{\mathrm{i} z}}\):
04

Write the formula for \(\mathrm{e}^{\mathrm{i} z}\) and \(\mathrm{e}^{\mathrm{i} \bar{z}}\)

Using Euler's formula, write the formula for \(\mathrm{e}^{\mathrm{i} z}\): \(\mathrm{e}^{\mathrm{i} z}=\cos(x)\cos(y) + i\cdot\sin(x)\cosh(y)\). Similarly, write the formula for \(\mathrm{e}^{\mathrm{i} \bar{z}}\): \(\mathrm{e}^{\mathrm{i} \bar{z}}=\cos(x)\cos(y) - i\cdot\sin(x)\cosh(y)\).
05

Find the complex conjugate of \(\mathrm{e}^{\mathrm{i} z}\)

The complex conjugate of \(\mathrm{e}^{\mathrm{i} z}\) is \(\overline{\mathrm{e}^{\mathrm{i} z}}=\cos(x)\cos(y) - i\cdot\sin(x)\cosh(y)\).
06

Equate the expressions and solve for 'z'

Equate the expressions from steps 1 and 2: \(\mathrm{e}^{\mathrm{i} \bar{z}} = \overline{\mathrm{e}^{\mathrm{i} z}}\), which gives \(\cos(x)\cos(y) - i\cdot\sin(x)\cosh(y)=\cos(x)\cos(y) - i\cdot\sin(x)\cosh(y)\). Since these two expressions are equal, this equation is satisfied for all \(z = x + iy \in \mathbb{C}\) where \(x\) and \(y\) are real numbers and \(y = n\pi\) for \(n \in \mathbb{Z}\) (because \(\sin(x) = 0\) for \(x=n\pi\), and \(\mathrm{e}^{\mathrm{i} z}\) and \(\mathrm{e}^{\mathrm{i} \bar{z}}\) must be real).

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Most popular questions from this chapter

F眉r welche \(x \in \mathbb{R}\) konvergieren die folgenden Potenzreihen? (a) \(\left(\sum_{n=1}^{\infty} \frac{(-1)^{n}\left(2^{n}+1\right)}{n}\left(x-\frac{1}{2}\right)^{n}\right)\) (b) \(\left(\sum_{n=0}^{\infty} \frac{1-(-2)^{-n-1} n !}{n !}(x-2)^{n}\right)\) (c) \(\left(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\left[\sqrt{n^{2}+n}-\sqrt{n^{2}+1}\right]^{n}(x+1)^{n}\right)\)

Gesucht ist eine Potenzreihendarstellung der Form \(\left(\sum_{n=0}^{\infty} a_{n} x^{n}\right)\) zu der Funktion $$ f(x)=\frac{\mathrm{e}^{x}}{1-x}, \quad x \in \mathbb{R} \backslash\\{1\\} $$ (a) Zeigen Sie \(a_{n}=\sum_{k=0}^{n} \frac{1}{k !}\). (b) F眉r welche \(x \in \mathbb{R}\) konvergiert die Potenzreihe?

Berechnen Sie mit dem Taschenrechner die Differenz \(\sin (\sinh (x))-\sinh (\sin (x))\) f眉r \(x \in\\{0.1,0.01,0.001\\}\) Erkl盲ren Sie diese Beobachtung, indem Sie das erste Glied der Potenzreihenentwicklung dieser Differenz um den Entwicklungspunkt 0 bestimmen.

Welche der folgenden Aussagen 眉ber eine Potenzreihe mit Entwicklungspunkt \(z_{0} \in \mathbb{C}\) und Konvergenzradius \(\rho\) sind richtig? (a) Die Potenzreihe konvergiert f眉r alle \(z \in \mathbb{C}\) mit \(\left|z-z_{0}\right|<\rho\) absolut. (b) Die Potenzreihe ist eine auf dem Konvergenzkreis beschr盲nkte Funktion. (c) Die Potenzreihe ist auf jedem Kreis mit Mittelpunkt \(z_{0}\) und Radius \(r<\rho\) eine beschr盲nkte Funktion. (d) Die Potenzreihe konvergiert f眉r kein \(z \in \mathbb{C}\) mit \(\left|z-z_{0}\right|=\rho\). (e) Konvergiert die Potenzreihe f眉r ein \(\hat{z} \in \mathbb{C}\) mit \(\left|\hat{z}-z_{0}\right|=\rho\) absolut, so gilt dies f眉r alle \(z \in \mathbb{C}\) mit \(\left|z-z_{0}\right|=\rho\).

Bestimmen Sie den Konvergenzradius und den Konvergenzkreis der folgenden Potenzreihen. (a) \(\left(\sum_{k=0}^{\infty} \frac{(k !)^{4}}{(4 k) !} z^{k}\right)\) (b) \(\left(\sum_{n=1}^{\infty} n^{n}(z-2)^{n}\right)\) (c) \(\left(\sum_{n=0}^{\infty} \frac{n+\mathrm{i}}{(\sqrt{2} \mathrm{i})^{n}}\left(\begin{array}{c}2 n \\ n\end{array}\right) z^{2 n}\right)\) (d) \(\left(\sum_{n=0}^{\infty} \frac{(2+\mathrm{i})^{n}-\mathrm{i}}{\mathrm{i}^{n}}(z+\mathrm{i})^{n}\right)\)
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