Exercise 1.3: Calculating with Complex Numbers

Considered numbers
in the complex plane

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 quantities 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:

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 complex quantity  $$z_4 = z_2^2 + z_3^2 = x_4 + {\rm j} \cdot y_4$$?

 $$x_4 \ =\$$ $$y_4 \ =\$$

3

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

 $$x_5 \ =\$$ $$y_5 \ =\$$

4

$$z_6$$  is the square root of  $$z_3$$.  Therefore $$z_6$$  has two solutions with the magnitude  $$|z_6| = 1$$.
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

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

 $$x_8 \ =\$$ $$y_8 \ =\$$

Solution

Solution

(1)  Correct are the solutions 1 and 2:

$2 \cdot z_1 + z_2 = 2 \cdot \cos(45^{ \circ}) - 2 \cdot {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \sin(45^{ \circ})- 2 \cdot \cos(45^{ \circ}) + 2\cdot {\rm j} \cdot\sin(45^{ \circ}) = 0.$
• 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)  The square of  $$z_2$$  has the magnitude  $$|z_2|^{2}$$  and the Phase  $$2 \cdot \phi_2$$:

$z_2^2 = 2^2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 270^{ \circ}}= 4 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 90^{ \circ}}=-4 \cdot {\rm j}.$
• Accordingly, the following applies to the square of  $$z_3$$:
$z_3^2 = (-{\rm j})^2 = -1.$
• Thus  $$x_4 =\underline{ –1}$$  and  $$y_4 = \underline{–4}.$$

(3)  By applying the division rule one obtains:

$z_5 = {1}/{z_2} = \frac{1}{2 \cdot{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 135^{ \circ}}}= 0.5 \cdot{\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 135^{ \circ}} = 0.5 \cdot \big[ \cos (- 135^{ \circ}) + {\rm j} \cdot \sin (- 135^{ \circ})\big]$
$\Rightarrow \ x_5 = - {\sqrt{2}}/{4}\hspace{0.15cm}\underline{= -0.354},\hspace{0.5cm} y_5 = x_5 \hspace{0.15cm}\underline{= -0.354}.$

(4)  The given relation for  $$z_6$$  can be transformed as follows:  $$z_6^2 = {z_3} = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 90^{ \circ}}.$$

• We can see that there are two possibilities for  $$z_6$$  that satisfy this equation:
$z_6 \hspace{0.1cm}{\rm (1.\hspace{0.1cm} solution)}\hspace{0.1cm} = \frac{z_2}{2} = 1 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}135^{ \circ}} \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{= 135^{ \circ}},$
$z_6 \hspace{0.1cm}{\rm (2.\hspace{0.1cm} solution)}\hspace{0.1cm} = {z_1} = 1 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}45^{ \circ}} \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{=-45^{ \circ}}.$

(5)  The complex quantity  $$z_2$$  in real part/imaginary part representation is:

$z_2 = x_2 + {\rm j} \cdot y_2 = -\sqrt{2} + {\rm j} \cdot\sqrt{2}.$
• This results in the following for the complex exponential function:
$z_7 = {\rm e}^{-\sqrt{2} + {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{2}}= {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2})\big].$
• Thus with  $${\rm e}^{-\sqrt{2} } = 0.243, \hspace{0.4cm} \cos (\sqrt{2}) = 0.156, \hspace{0.4cm} \sin (\sqrt{2}) = 0.988$$  one obtains:
$z_7 = 0.243 \cdot \left( 0.156 + {\rm j} \cdot 0.988\right) \hspace{0.15cm}\underline{= 0.038 + {\rm j} \cdot 0.24}.$

(6)  Starting from the result of subtask  (4)  one obtains for $$z_8$$:

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