Aufgaben:Exercise 1.3: Calculating with Complex Numbers

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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:

• This exercise belongs to the chapter Calculating with Complex Numbers.
• The topic of this task is also covered in the (German language) learning video
       Rechnen mit komplexen Zahlen   ⇒   "Arithmetic operations involving complex numbers".

Questions

1 Which of the following equations are true?

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

2 What is the value of the complex quantity  [math]\displaystyle{ z_4 = z_2^2 + z_3^2 = x_4 + {\rm j} \cdot y_4 }[/math]?

[math]\displaystyle{ x_4 \ =\ }[/math]

[math]\displaystyle{ y_4 \ =\ }[/math]

3 Calculate the complex quantity  [math]\displaystyle{ z_5 = 1/z_2 = x_5 + {\rm j} \cdot y_5 }[/math].

[math]\displaystyle{ x_5 \ =\ }[/math]

[math]\displaystyle{ y_5 \ =\ }[/math]

4 [math]\displaystyle{ z_6 }[/math]  is the square root of  [math]\displaystyle{ z_3 }[/math].  Therefore [math]\displaystyle{ z_6 }[/math]  has two solutions with the magnitude  [math]\displaystyle{ |z_6| = 1 }[/math].
Give the two possible phase angles of  [math]\displaystyle{ z_6 }[/math] .

[math]\displaystyle{ \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} =\ }[/math] $\ \text{deg}$

[math]\displaystyle{ \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} =\ }[/math] $\ \text{deg}$

5 Calculate  [math]\displaystyle{ z_7 = {\rm e}^{z_2} = x_7 + {\rm j} \cdot y_7 }[/math].

[math]\displaystyle{ x_7 \ =\ }[/math]

[math]\displaystyle{ y_7 \ =\ }[/math]

6 Calculate the complex quantity  [math]\displaystyle{ z_8 = {\rm e}^{z_2} + {\rm e}^{z_2^{\ast}} = x_8 + {\rm j}\cdot y_8 }[/math] .

[math]\displaystyle{ x_8 \ =\ }[/math]

[math]\displaystyle{ y_8 \ =\ }[/math]

Solution

Solution

(1)  Correct are the solutions 1 and 2:

• The following applies with  Euler's theorem: 

[math]\displaystyle{ 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. }[/math]

• The second option is also correct, because

[math]\displaystyle{ 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. }[/math]

• In contrast, the third option is wrong. The division of  [math]\displaystyle{ z_1 }[/math] and [math]\displaystyle{ z_2 }[/math]  yields: 

[math]\displaystyle{ \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. }[/math]

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

(2)  The square of  [math]\displaystyle{ z_2 }[/math]  has the magnitude  [math]\displaystyle{ |z_2|^{2} }[/math]  and the Phase  [math]\displaystyle{ 2 \cdot \phi_2 }[/math]: 

[math]\displaystyle{ 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}. }[/math]

• Accordingly, the following applies to the square of  [math]\displaystyle{ z_3 }[/math]:

[math]\displaystyle{ z_3^2 = (-{\rm j})^2 = -1. }[/math]

• Thus  [math]\displaystyle{ x_4 =\underline{ –1} }[/math]  and  [math]\displaystyle{ y_4 = \underline{–4}. }[/math]

(3)  By applying the division rule one obtains: 

[math]\displaystyle{ 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] }[/math]

[math]\displaystyle{ \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}. }[/math]

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

• We can see that there are two possibilities for  [math]\displaystyle{ z_6 }[/math]  that satisfy this equation:  

[math]\displaystyle{ 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}}, }[/math]

[math]\displaystyle{ 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}}. }[/math]

(5)  The complex quantity  [math]\displaystyle{ z_2 }[/math]  in real part/imaginary part representation is: 

[math]\displaystyle{ z_2 = x_2 + {\rm j} \cdot y_2 = -\sqrt{2} + {\rm j} \cdot\sqrt{2}. }[/math]

• This results in the following for the complex exponential function:

[math]\displaystyle{ 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]. }[/math]

• Thus with  [math]\displaystyle{ {\rm e}^{-\sqrt{2} } = 0.243, \hspace{0.4cm} \cos (\sqrt{2}) = 0.156, \hspace{0.4cm} \sin (\sqrt{2}) = 0.988 }[/math]  one obtains: 

[math]\displaystyle{ 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}. }[/math]

(6)  Starting from the result of subtask  (4)  one obtains for [math]\displaystyle{ z_8 }[/math]: 

[math]\displaystyle{ 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}. }[/math]