Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Given two complex numbers, [tex]\( z_1 = -i \)[/tex] and [tex]\( z_2 = 2 + i \sqrt{3} \)[/tex].
### Part (a)
First, we will express [tex]\( z_1^2 \)[/tex] and [tex]\( \overline{z_2} \)[/tex] in the form [tex]\( a + bi \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are real numbers.
#### 1. Calculate [tex]\( z_1^2 \)[/tex]:
[tex]\[ z_1 = -i \][/tex]
[tex]\[ z_1^2 = (-i)^2 \][/tex]
Knowing that [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ z_1^2 = (-i)^2 = (-i)(-i) = i^2 = -1 \][/tex]
Therefore, in the form [tex]\( a + bi \)[/tex], we have:
[tex]\[ z_1^2 = -1 + 0i \][/tex]
Thus, [tex]\( a = -1 \)[/tex] and [tex]\( b = 0 \)[/tex].
#### 2. Calculate the conjugate of [tex]\( z_2 \)[/tex]:
[tex]\[ z_2 = 2 + i \sqrt{3} \][/tex]
The conjugate of [tex]\( z_2 \)[/tex] is obtained by changing the sign of the imaginary part:
[tex]\[ \overline{z_2} = 2 - i \sqrt{3} \][/tex]
Expressing this in the form [tex]\( a + bi \)[/tex]:
[tex]\[ \overline{z_2} = 2 + (-\sqrt{3})i \][/tex]
Thus, [tex]\( a = 2 \)[/tex] and [tex]\( b = -\sqrt{3} \)[/tex].
So, for part (a), we have:
- [tex]\( z_1^2 \)[/tex] in the form [tex]\( a + bi \)[/tex] is [tex]\( -1 + 0i \)[/tex].
- [tex]\( \overline{z_2} \)[/tex] in the form [tex]\( a + bi \)[/tex] is [tex]\( 2 - \sqrt{3}i \)[/tex].
### Part (b)
Using the results from part (a), we need to find [tex]\( w \)[/tex] given by:
[tex]\[ w = \frac{z_1^2 + \overline{z_2}}{z_1} \][/tex]
Substitute [tex]\( z_1^2 \)[/tex] and [tex]\( \overline{z_2} \)[/tex] into the equation:
[tex]\[ w = \frac{(-1 + 0i) + (2 - \sqrt{3}i)}{-i} \][/tex]
First, combine the numerator:
[tex]\[ z_1^2 + \overline{z_2} = (-1 + 0i) + (2 - \sqrt{3}i) = 1 - \sqrt{3}i \][/tex]
Next, divide by [tex]\( z_1 \)[/tex]:
[tex]\[ w = \frac{1 - \sqrt{3}i}{-i} \][/tex]
To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ w = \frac{(1 - \sqrt{3}i)(i)}{-i(i)} \][/tex]
[tex]\[ w = \frac{i - i^2 \sqrt{3}}{-i^2} \][/tex]
[tex]\[ w = \frac{i - (-1) \sqrt{3}}{1} \][/tex]
Since [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ w = i + \sqrt{3} \][/tex]
Express [tex]\( w \)[/tex] as:
[tex]\[ w = \sqrt{3} + i \][/tex]
#### Calculate the magnitude [tex]\( |w| \)[/tex]:
The magnitude of a complex number [tex]\( a + bi \)[/tex] is given by:
[tex]\[ |w| = \sqrt{a^2 + b^2} \][/tex]
For [tex]\( w = \sqrt{3} + i \)[/tex]:
[tex]\[ |w| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \][/tex]
#### Calculate the argument [tex]\( \arg(w) \)[/tex]:
The argument of a complex number [tex]\( a + bi \)[/tex] is given by:
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{b}{a} \right) \][/tex]
For [tex]\( w = \sqrt{3} + i \)[/tex]:
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \][/tex]
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \text{ radians} \][/tex]
In conclusion:
- [tex]\( w = \sqrt{3} + i \)[/tex]
- The magnitude [tex]\( |w| = 2 \)[/tex]
- The argument [tex]\( \arg(w) = \frac{\pi}{6} \)[/tex] radians
### Part (a)
First, we will express [tex]\( z_1^2 \)[/tex] and [tex]\( \overline{z_2} \)[/tex] in the form [tex]\( a + bi \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are real numbers.
#### 1. Calculate [tex]\( z_1^2 \)[/tex]:
[tex]\[ z_1 = -i \][/tex]
[tex]\[ z_1^2 = (-i)^2 \][/tex]
Knowing that [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ z_1^2 = (-i)^2 = (-i)(-i) = i^2 = -1 \][/tex]
Therefore, in the form [tex]\( a + bi \)[/tex], we have:
[tex]\[ z_1^2 = -1 + 0i \][/tex]
Thus, [tex]\( a = -1 \)[/tex] and [tex]\( b = 0 \)[/tex].
#### 2. Calculate the conjugate of [tex]\( z_2 \)[/tex]:
[tex]\[ z_2 = 2 + i \sqrt{3} \][/tex]
The conjugate of [tex]\( z_2 \)[/tex] is obtained by changing the sign of the imaginary part:
[tex]\[ \overline{z_2} = 2 - i \sqrt{3} \][/tex]
Expressing this in the form [tex]\( a + bi \)[/tex]:
[tex]\[ \overline{z_2} = 2 + (-\sqrt{3})i \][/tex]
Thus, [tex]\( a = 2 \)[/tex] and [tex]\( b = -\sqrt{3} \)[/tex].
So, for part (a), we have:
- [tex]\( z_1^2 \)[/tex] in the form [tex]\( a + bi \)[/tex] is [tex]\( -1 + 0i \)[/tex].
- [tex]\( \overline{z_2} \)[/tex] in the form [tex]\( a + bi \)[/tex] is [tex]\( 2 - \sqrt{3}i \)[/tex].
### Part (b)
Using the results from part (a), we need to find [tex]\( w \)[/tex] given by:
[tex]\[ w = \frac{z_1^2 + \overline{z_2}}{z_1} \][/tex]
Substitute [tex]\( z_1^2 \)[/tex] and [tex]\( \overline{z_2} \)[/tex] into the equation:
[tex]\[ w = \frac{(-1 + 0i) + (2 - \sqrt{3}i)}{-i} \][/tex]
First, combine the numerator:
[tex]\[ z_1^2 + \overline{z_2} = (-1 + 0i) + (2 - \sqrt{3}i) = 1 - \sqrt{3}i \][/tex]
Next, divide by [tex]\( z_1 \)[/tex]:
[tex]\[ w = \frac{1 - \sqrt{3}i}{-i} \][/tex]
To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ w = \frac{(1 - \sqrt{3}i)(i)}{-i(i)} \][/tex]
[tex]\[ w = \frac{i - i^2 \sqrt{3}}{-i^2} \][/tex]
[tex]\[ w = \frac{i - (-1) \sqrt{3}}{1} \][/tex]
Since [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ w = i + \sqrt{3} \][/tex]
Express [tex]\( w \)[/tex] as:
[tex]\[ w = \sqrt{3} + i \][/tex]
#### Calculate the magnitude [tex]\( |w| \)[/tex]:
The magnitude of a complex number [tex]\( a + bi \)[/tex] is given by:
[tex]\[ |w| = \sqrt{a^2 + b^2} \][/tex]
For [tex]\( w = \sqrt{3} + i \)[/tex]:
[tex]\[ |w| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \][/tex]
#### Calculate the argument [tex]\( \arg(w) \)[/tex]:
The argument of a complex number [tex]\( a + bi \)[/tex] is given by:
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{b}{a} \right) \][/tex]
For [tex]\( w = \sqrt{3} + i \)[/tex]:
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \][/tex]
[tex]\[ \arg(w) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \text{ radians} \][/tex]
In conclusion:
- [tex]\( w = \sqrt{3} + i \)[/tex]
- The magnitude [tex]\( |w| = 2 \)[/tex]
- The argument [tex]\( \arg(w) = \frac{\pi}{6} \)[/tex] radians
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
