Exercise 1.3: Calculating with Complex Numbers

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Betrachtete Zahlen in der komplexen Ebene

The diagram to the right shows some points in the complex plane, namely

$$z_1 = {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 45^{ \circ}},$$

$$z_2 = 2 \cdot{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}135^{ \circ}},$$

$$z_3 = -{\rm j} .$$

In the course of this task, the following complex values will be considered:

$$z_4 = z_2^2 + z_3^2,$$

$$z_5 = 1/z_2,$$

$$z_6 = \sqrt{z_3},$$

$$z_7 = {\rm e}^{\hspace{0.05cm}z_2},$$

$$z_8 = {\rm e}^{\hspace{0.05cm}z_2} + {\rm e}^{\hspace{0.05cm}z_2^{\star}}.$$

Notes:

• This exercise belongs to the chapter Calculating With Complex Numbers.
• The topic is also covered in the teaching video Rechnen mit komplexen Zahlen .

Questions

1

Which of the following equations are true?

	$2 \cdot z_1 + z_2 =0.$
	$z_1^{\ast} \cdot z_2 +2=0.$
	$(z_1/z_2) \cdot z_3$ is purely real.

2

What is the value of the random variable  $z_4 = z_2^2 + z_3^2 = x_4 + {\rm j} \cdot y_4$?

$x_4 \ =\ $

$y_4 \ =\ $

3

Calculate the complex value  $z_5 = 1/z_2 = x_5 + {\rm j} \cdot y_5$.

$x_5 \ =\ $

$y_5 \ =\ $

4

$z_6$  is the square root of  $z_3$ . Therefore $z_6$  has two solutions with the absolute value  $|z_6| = 1$.
Give the two possible phase angles of  $z_6$ .

$\phi_6 \ ({\rm between\hspace{0.1cm} 0^{\circ} \hspace{0.1cm}and \hspace{0.1cm} +\hspace{-0.15cm}180^{\circ} \hspace{0.1cm}deg}) \hspace{0.2cm} =\ $ $\ \text{deg}$

$\phi_6 \ ({\rm between\hspace{0.1cm} - \hspace{-0.15cm}180^{\circ} \hspace{0.1cm}and \hspace{0.1cm} 0^{\circ} \hspace{0.1cm}deg}) \hspace{0.2cm} =\ $ $\ \text{deg}$

5

Calculate  $z_7 = {\rm e}^{z_2} = x_7 + {\rm j} \cdot y_7$.

$x_7 \ =\ $

$y_7 \ =\ $

6

Compute the complex value  $z_8 = {\rm e}^{z_2} + {\rm e}^{z_2^{\ast}} = x_8 + {\rm j}\cdot y_8$ .

$x_8 \ =\ $

$y_8 \ =\ $

Solution

Solution

(1)  Correct are the solutions 1 and 2:

• The following applies with  Euler's theorem: 

$$2 \cdot z_1 + z_2 = 2 \cdot \cos(45^{ \circ}) - 2 \cdot {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \sin(45^{ \circ})- 2 \cdot \cos(45^{ \circ}) + 2\cdot {\rm j} \cdot\sin(45^{ \circ}) = 0.$$

• The second option is also correct, because

$$z_1^{\star} \cdot z_2 = 1 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 45^{ \circ}} \cdot 2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 135^{ \circ}} = 2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 180^{ \circ}}= -2.$$

• In contrast, the third option is wrong. The division of  $z_1$ and $z_2$  yields: 

$$\frac{z_1}{z_2} = \frac{{\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 45^{ \circ}}}{2 \cdot{\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}135^{ \circ}}} = 0.5 \cdot{\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 180^{ \circ}}= -0.5.$$

• The multiplication by  $z_3 = -{\rm j}$  leads to the result  ${\rm j}/2$, i.e. to a purely imaginary quantity.

(2)  Das Quadrat von  $z_2$  hat den Betrag  $|z_2|^{2}$  und die Phase  $2 \cdot \phi_2$: 

$$z_2^2 = 2^2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 270^{ \circ}}= 4 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 90^{ \circ}}=-4 \cdot {\rm j}.$$

• Entsprechend gilt für das Quadrat von  $z_3$: 

$$z_3^2 = (-{\rm j})^2 = -1.$$

• Somit ist  $x_4 =\underline{ –1}$  und  $y_4 = \underline{–4}.$

(3)  Durch Anwendung der Divisionsregel erhält man: 

$$z_5 = {1}/{z_2} = \frac{1}{2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 135^{ \circ}}}= 0.5 \cdot{\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 135^{ \circ}} = 0.5 \cdot \big[ \cos (- 135^{ \circ}) + {\rm j} \cdot \sin (- 135^{ \circ})\big]$$

$$\Rightarrow \ x_5 = - {\sqrt{2}}/{4}\hspace{0.15cm}\underline{= -0.354},\hspace{0.5cm} y_5 = x_5 \hspace{0.15cm}\underline{= -0.354}.$$

(4)  Die angegeben Beziehung für  $z_6$  kann wie folgt umgeformt werden:  $z_6^2 = {z_3} = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 90^{ \circ}}.$

• Man erkennt, dass es zwei Möglichkeiten für  $z_6$  gibt, die diese Gleichung erfüllen: 

$$z_6 \hspace{0.1cm}{\rm (1.\hspace{0.1cm} L\ddot{o}sung)}\hspace{0.1cm} = \frac{z_2}{2} = 1 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}135^{ \circ}} \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{= 135^{ \circ}},$$

$$z_6 \hspace{0.1cm}{\rm (2.\hspace{0.1cm} L \ddot{o}sung)}\hspace{0.1cm} = {z_1} = 1 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}45^{ \circ}} \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{=-45^{ \circ}}.$$

(5)  Die komplexe Größe  $z_2$  lautet in Realteil/Imaginärteildarstellung: 

$$z_2 = x_2 + {\rm j} \cdot y_2 = -\sqrt{2} + {\rm j} \cdot\sqrt{2}.$$

• Damit ergibt sich für die komplexe Exponentialfunktion:

$$z_7 = {\rm e}^{-\sqrt{2} + {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{2}}= {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2})\big].$$

• Mit  ${\rm e}^{-\sqrt{2} } = 0.243, \hspace{0.4cm} \cos (\sqrt{2}) = 0.156, \hspace{0.4cm} \sin (\sqrt{2}) = 0.988$  erhält man somit: 

$$z_7 = 0.243 \cdot \left( 0.156 + {\rm j} \cdot 0.988\right) \hspace{0.15cm}\underline{= 0.038 + {\rm j} \cdot 0.24}.$$

(6)  Ausgehend vom Ergebnis der Teilaufgabe  (4)  erhält man für $z_8$: 

$$z_8 = {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2}) + \cos (\sqrt{2}) - {\rm j} \cdot \sin (\sqrt{2})\big] = 2 \cdot {\rm e}^{-\sqrt{2} } \cdot \cos (\sqrt{2}) = 2 \cdot x_7 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} x_8 \hspace{0.15cm}\underline{= 0.076}, \hspace{0.4cm}y_8\hspace{0.15cm}\underline{ = 0}.$$