Difference between revisions of "Aufgaben:Exercise 1.3: Calculating with Complex Numbers"
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:$$z_3 = -{\rm j} .$$ | :$$z_3 = -{\rm j} .$$ | ||
| − | In the course of this task, the following complex | + | In the course of this task, the following complex quantities will be considered: |
:$$z_4 = z_2^2 + z_3^2,$$ | :$$z_4 = z_2^2 + z_3^2,$$ | ||
:$$z_5 = 1/z_2,$$ | :$$z_5 = 1/z_2,$$ | ||
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| − | {Calculate the complex | + | {Calculate the complex quantity <math>z_5 = 1/z_2 = x_5 + {\rm j} \cdot y_5</math>. |
|type="{}"} | |type="{}"} | ||
<math> x_5 \ =\ </math> { -0.36--0.35 } | <math> x_5 \ =\ </math> { -0.36--0.35 } | ||
<math> y_5 \ =\ </math> { -0.36--0.35 } | <math> y_5 \ =\ </math> { -0.36--0.35 } | ||
| − | {<math>z_6</math> is the square root of <math>z_3</math> . Therefore <math>z_6</math> has two solutions with the | + | {<math>z_6</math> is the square root of <math>z_3</math> . Therefore <math>z_6</math> has two solutions with the magnitude <math>|z_6| = 1</math>. <br>Give the two possible phase angles of <math>z_6</math> . |
|type="{}"} | |type="{}"} | ||
<math> \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> { 133-137 } $\ \text{deg}$ | <math> \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> { 133-137 } $\ \text{deg}$ | ||
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| − | {Compute the complex | + | {Compute the complex quantity <math>z_8 = {\rm e}^{z_2} + {\rm e}^{z_2^{\ast}} = x_8 + {\rm j}\cdot y_8</math> . |
|type="{}"} | |type="{}"} | ||
<math> x_8 \ =\ </math> { 0.07-0.08 } | <math> x_8 \ =\ </math> { 0.07-0.08 } | ||
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| − | '''(2)''' | + | '''(2)''' The square of <math>z_2</math> has the magnitude <math>|z_2|^{2}</math> and the Phase <math>2 \cdot \phi_2</math>: |
::<math>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> | ::<math>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>z_3</math>: : |
::<math>z_3^2 = (-{\rm j})^2 = -1.</math> | ::<math>z_3^2 = (-{\rm j})^2 = -1.</math> | ||
| − | * | + | *Thus <math>x_4 =\underline{ –1}</math> and <math>y_4 = \underline{–4}.</math> |
| − | '''(3)''' | + | '''(3)''' By applying the division rule one obtains: |
::<math>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^{ | ::<math>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^{ | ||
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| − | '''(4)''' | + | '''(4)''' The given relation for <math>z_6</math> can be transformed as follows: <math>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>z_6</math> that satisfy this equation: |
| − | ::<math>z_6 \hspace{0.1cm}{\rm (1.\hspace{0.1cm} | + | ::<math>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^{ | \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{= 135^{ | ||
\circ}}, </math> | \circ}}, </math> | ||
| − | ::<math>z_6 \hspace{0.1cm}{\rm (2.\hspace{0.1cm} | + | ::<math>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^{ | \hspace{0.2cm}\Rightarrow \hspace{0.2cm} \phi_6 \hspace{0.15cm}\underline{=-45^{ | ||
\circ}}.</math> | \circ}}.</math> | ||
| − | '''(5)''' | + | '''(5)''' The complex quantity <math>z_2</math> in real part/imaginary part representation is: |
::<math>z_2 = x_2 + {\rm j} \cdot y_2 = -\sqrt{2} + {\rm j} \cdot\sqrt{2}.</math> | ::<math>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>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> | ::<math>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>{\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>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> | ::<math>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> | ||
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| − | '''(6)''' | + | '''(6)''' Starting from the result of subtask '''(4)''' one obtains for <math>z_8</math>: |
::<math>z_8 = {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2}) + \cos (\sqrt{2}) - {\rm j} \cdot \sin | ::<math>z_8 = {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2}) + \cos (\sqrt{2}) - {\rm j} \cdot \sin | ||
Revision as of 15:16, 25 December 2020
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 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 is also covered in the teaching video Rechnen mit komplexen Zahlen .
Questions
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) 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 \(z_8\):
- \[z_8 = {\rm e}^{-\sqrt{2} } \cdot \big[ \cos (\sqrt{2}) + {\rm j} \cdot \sin (\sqrt{2}) + \cos (\sqrt{2}) - {\rm j} \cdot \sin (\sqrt{2})\big] = 2 \cdot {\rm e}^{-\sqrt{2} } \cdot \cos (\sqrt{2}) = 2 \cdot x_7 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} x_8 \hspace{0.15cm}\underline{= 0.076}, \hspace{0.4cm}y_8\hspace{0.15cm}\underline{ = 0}.\]
