Answer :
Let's solve the problem step by step.
First, we are given the complex numbers [tex]\( w \)[/tex] and [tex]\( z \)[/tex] in polar form:
[tex]\[ w = \sqrt{2}\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) \][/tex]
[tex]\[ z = 2\left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right) \][/tex]
To perform the subtraction [tex]\( w - z \)[/tex], we need to convert these numbers to rectangular form.
1. Converting [tex]\( w \)[/tex] to rectangular form:
- Magnitude [tex]\( w_r = \sqrt{2} \)[/tex]
- Angle [tex]\( w_\theta = \frac{\pi}{4} \)[/tex]
Using the rectangular form formula:
[tex]\[ w_{\text{re}} = w_r \cos(w_\theta) = \sqrt{2} \cos\left(\frac{\pi}{4}\right) \][/tex]
[tex]\[ w_{\text{im}} = w_r \sin(w_\theta) = \sqrt{2} \sin\left(\frac{\pi}{4}\right) \][/tex]
We know:
[tex]\[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \][/tex]
Therefore:
[tex]\[ w_{\text{re}} = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \][/tex]
[tex]\[ w_{\text{im}} = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \][/tex]
2. Converting [tex]\( z \)[/tex] to rectangular form:
- Magnitude [tex]\( z_r = 2 \)[/tex]
- Angle [tex]\( z_\theta = \frac{\pi}{2} \)[/tex]
Using the rectangular form formula:
[tex]\[ z_{\text{re}} = z_r \cos(z_\theta) = 2 \cos\left(\frac{\pi}{2}\right) \][/tex]
[tex]\[ z_{\text{im}} = z_r \sin(z_\theta) = 2 \sin\left(\frac{\pi}{2}\right) \][/tex]
We know:
[tex]\[ \cos\left(\frac{\pi}{2}\right) = 0 \][/tex]
[tex]\[ \sin\left(\frac{\pi}{2}\right) = 1 \][/tex]
Therefore:
[tex]\[ z_{\text{re}} = 2 \cdot 0 = 0 \][/tex]
[tex]\[ z_{\text{im}} = 2 \cdot 1 = 2 \][/tex]
3. Subtracting [tex]\( w \)[/tex] and [tex]\( z \)[/tex] in rectangular form:
[tex]\[ w - z = (w_{\text{re}} + i w_{\text{im}}) - (z_{\text{re}} + i z_{\text{im}}) \][/tex]
[tex]\[ w - z = (1 + i \cdot 1) - (0 + i \cdot 2) \][/tex]
[tex]\[ w - z = (1 - 0) + i(1 - 2) \][/tex]
[tex]\[ w - z = 1 - i \][/tex]
4. Converting the result back to polar form:
- The magnitude [tex]\( r \)[/tex] is given by:
[tex]\[ r = \sqrt{(\text{Re})^2 + (\text{Im})^2} \][/tex]
[tex]\[ r = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
- The angle [tex]\( \theta \)[/tex] is given by:
[tex]\[ \theta = \tan^{-2}\left(\frac{\text{Im}}{\text{Re}}\right) = \tan^{-1}\left(\frac{-1}{1}\right) = \tan^{-1}(-1) \][/tex]
We know:
[tex]\[ \tan^{-1}(-1) = -\frac{\pi}{4} \][/tex]
To bring the angle within the standard range of [tex]\( [0, 2\pi) \)[/tex], we add [tex]\( 2\pi \)[/tex]:
[tex]\[ \theta = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4} \][/tex]
So, [tex]\( w - z \)[/tex] expressed in polar form is:
[tex]\[ w - z = \sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right) \][/tex]
Therefore, the correct option is:
[tex]\[ \sqrt{2}\left(\cos \left(\frac{7 \pi}{4}\right)+i \sin \left(\frac{7 \pi}{4}\right)\right) \][/tex]
First, we are given the complex numbers [tex]\( w \)[/tex] and [tex]\( z \)[/tex] in polar form:
[tex]\[ w = \sqrt{2}\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) \][/tex]
[tex]\[ z = 2\left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right) \][/tex]
To perform the subtraction [tex]\( w - z \)[/tex], we need to convert these numbers to rectangular form.
1. Converting [tex]\( w \)[/tex] to rectangular form:
- Magnitude [tex]\( w_r = \sqrt{2} \)[/tex]
- Angle [tex]\( w_\theta = \frac{\pi}{4} \)[/tex]
Using the rectangular form formula:
[tex]\[ w_{\text{re}} = w_r \cos(w_\theta) = \sqrt{2} \cos\left(\frac{\pi}{4}\right) \][/tex]
[tex]\[ w_{\text{im}} = w_r \sin(w_\theta) = \sqrt{2} \sin\left(\frac{\pi}{4}\right) \][/tex]
We know:
[tex]\[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \][/tex]
Therefore:
[tex]\[ w_{\text{re}} = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \][/tex]
[tex]\[ w_{\text{im}} = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \][/tex]
2. Converting [tex]\( z \)[/tex] to rectangular form:
- Magnitude [tex]\( z_r = 2 \)[/tex]
- Angle [tex]\( z_\theta = \frac{\pi}{2} \)[/tex]
Using the rectangular form formula:
[tex]\[ z_{\text{re}} = z_r \cos(z_\theta) = 2 \cos\left(\frac{\pi}{2}\right) \][/tex]
[tex]\[ z_{\text{im}} = z_r \sin(z_\theta) = 2 \sin\left(\frac{\pi}{2}\right) \][/tex]
We know:
[tex]\[ \cos\left(\frac{\pi}{2}\right) = 0 \][/tex]
[tex]\[ \sin\left(\frac{\pi}{2}\right) = 1 \][/tex]
Therefore:
[tex]\[ z_{\text{re}} = 2 \cdot 0 = 0 \][/tex]
[tex]\[ z_{\text{im}} = 2 \cdot 1 = 2 \][/tex]
3. Subtracting [tex]\( w \)[/tex] and [tex]\( z \)[/tex] in rectangular form:
[tex]\[ w - z = (w_{\text{re}} + i w_{\text{im}}) - (z_{\text{re}} + i z_{\text{im}}) \][/tex]
[tex]\[ w - z = (1 + i \cdot 1) - (0 + i \cdot 2) \][/tex]
[tex]\[ w - z = (1 - 0) + i(1 - 2) \][/tex]
[tex]\[ w - z = 1 - i \][/tex]
4. Converting the result back to polar form:
- The magnitude [tex]\( r \)[/tex] is given by:
[tex]\[ r = \sqrt{(\text{Re})^2 + (\text{Im})^2} \][/tex]
[tex]\[ r = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
- The angle [tex]\( \theta \)[/tex] is given by:
[tex]\[ \theta = \tan^{-2}\left(\frac{\text{Im}}{\text{Re}}\right) = \tan^{-1}\left(\frac{-1}{1}\right) = \tan^{-1}(-1) \][/tex]
We know:
[tex]\[ \tan^{-1}(-1) = -\frac{\pi}{4} \][/tex]
To bring the angle within the standard range of [tex]\( [0, 2\pi) \)[/tex], we add [tex]\( 2\pi \)[/tex]:
[tex]\[ \theta = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4} \][/tex]
So, [tex]\( w - z \)[/tex] expressed in polar form is:
[tex]\[ w - z = \sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right) \][/tex]
Therefore, the correct option is:
[tex]\[ \sqrt{2}\left(\cos \left(\frac{7 \pi}{4}\right)+i \sin \left(\frac{7 \pi}{4}\right)\right) \][/tex]
