Answer :
Certainly! Let's go through each part of the question step by step.
### Question 1
Solve [tex]\( z^3 - 2z - 4 = 0 \)[/tex] for the set of complex numbers.
#### Solution:
Given the hint [tex]\( z = 2 \)[/tex] is a solution to this equation, we need to find all roots of the polynomial.
The polynomial is:
[tex]\[ z^3 - 2z - 4 = 0 \][/tex]
We already know one root, [tex]\( z = 2 \)[/tex]. We can factor this polynomial using synthetic division or other methods, but the roots for the polynomial [tex]\( z^3 - 2z - 4 = 0 \)[/tex] are:
[tex]\[ z = 2, \, z = -1 - i, \, z = -1 + i \][/tex]
So, the solutions to the equation [tex]\( z^3 - 2z - 4 = 0 \)[/tex] are [tex]\( \{2, -1 - i, -1 + i\} \)[/tex].
### Question 2
Let [tex]\( z_1 = 1 + i \)[/tex], [tex]\( z_2 = 2 + i \)[/tex], and [tex]\( z_3 = 3 - i \)[/tex]. Calculate:
#### (a) [tex]\( z_1 z_2 \)[/tex]
To find the product of [tex]\( z_1 \)[/tex] and [tex]\( z_2 \)[/tex]:
[tex]\[ z_1 z_2 = (1+i)(2+i) \][/tex]
Expanding the product:
[tex]\[ (1+i)(2+i) = 1 \cdot 2 + 1 \cdot i + i \cdot 2 + i \cdot i \][/tex]
[tex]\[ = 2 + i + 2i + i^2 \][/tex]
[tex]\[ = 2 + 3i + (-1) \][/tex]
[tex]\[ = 1 + 3i \][/tex]
So, [tex]\( z_1 z_2 = 1 + 3i \)[/tex].
#### (b) [tex]\( |z_3| \)[/tex]
To find the magnitude of [tex]\( z_3 \)[/tex]:
[tex]\[ z_3 = 3 - i \][/tex]
The magnitude is given by:
[tex]\[ |z_3| = \sqrt{(\text{Re}(z_3))^2 + (\text{Im}(z_3))^2} \][/tex]
[tex]\[ = \sqrt{3^2 + (-1)^2} \][/tex]
[tex]\[ = \sqrt{9 + 1} \][/tex]
[tex]\[ = \sqrt{10} \][/tex]
[tex]\[ \approx 3.162 \][/tex]
So, [tex]\( |z_3| \approx 3.162 \)[/tex].
#### (c) [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) \)[/tex]
To find the imaginary part of [tex]\( \frac{z_1}{z_2} \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ z_2 = 2 + i \][/tex]
[tex]\[ \frac{z_1}{z_2} = \frac{1 + i}{2 + i} \][/tex]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{1 + i}{2 + i} \cdot \frac{2 - i}{2 - i} \][/tex]
[tex]\[ = \frac{(1+i)(2-i)}{(2+i)(2-i)} \][/tex]
[tex]\[ = \frac{2 + i - 2i - i^2}{4 + 1} \][/tex]
[tex]\[ = \frac{2 - i - (-1)}{5} \][/tex]
[tex]\[ = \frac{3 - i}{5} \][/tex]
[tex]\[ = \frac{3}{5} - \frac{i}{5} \][/tex]
[tex]\[ = 0.6 - 0.2i \][/tex]
The imaginary part is:
[tex]\[ \operatorname{Im}\left(\frac{z_1}{z_2}\right) = -0.2 \][/tex]
So, [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) = 0.2 \)[/tex].
#### (d) [tex]\( \operatorname{Arg}(\bar{z}_1) \)[/tex]
To find the argument of the conjugate of [tex]\( z_1 \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ \bar{z}_1 = 1 - i \][/tex]
The argument is the angle the complex number makes with the positive real axis.
For [tex]\( 1 - i \)[/tex], it is in the fourth quadrant:
[tex]\[ \theta = \arctan\left(\frac{-1}{1}\right) \][/tex]
[tex]\[ = \arctan(-1) \][/tex]
[tex]\[ = -\frac{\pi}{4} \][/tex]
So, [tex]\( \operatorname{Arg}(\bar{z}_1) = -\frac{\pi}{4} \approx -0.785 \)[/tex].
#### (e) [tex]\( z_1 + 2 z_3 \)[/tex]
To find [tex]\( z_1 + 2z_3 \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ z_3 = 3 - i \][/tex]
[tex]\[ z_1 + 2z_3 = (1 + i) + 2(3 - i) \][/tex]
[tex]\[ = 1 + i + 6 - 2i \][/tex]
[tex]\[ = 7 - i \][/tex]
So, [tex]\( z_1 + 2z_3 = 7 - i \)[/tex].
Hence:
1. [tex]\[ \text{Solutions to } z^3 - 2z - 4 = 0: \{2, -1 - i, -1 + i\} \][/tex]
2. [tex]\( z_1 z_2 = 1 + 3i \)[/tex]
3. [tex]\( |z_3| \approx 3.162 \)[/tex]
4. [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) = 0.2 \)[/tex]
5. [tex]\( \operatorname{Arg}(\bar{z}_1) = -\frac{\pi}{4} \approx -0.785 \)[/tex]
6. [tex]\( z_1 + 2z_3 = 7 - 1i \)[/tex]
### Question 1
Solve [tex]\( z^3 - 2z - 4 = 0 \)[/tex] for the set of complex numbers.
#### Solution:
Given the hint [tex]\( z = 2 \)[/tex] is a solution to this equation, we need to find all roots of the polynomial.
The polynomial is:
[tex]\[ z^3 - 2z - 4 = 0 \][/tex]
We already know one root, [tex]\( z = 2 \)[/tex]. We can factor this polynomial using synthetic division or other methods, but the roots for the polynomial [tex]\( z^3 - 2z - 4 = 0 \)[/tex] are:
[tex]\[ z = 2, \, z = -1 - i, \, z = -1 + i \][/tex]
So, the solutions to the equation [tex]\( z^3 - 2z - 4 = 0 \)[/tex] are [tex]\( \{2, -1 - i, -1 + i\} \)[/tex].
### Question 2
Let [tex]\( z_1 = 1 + i \)[/tex], [tex]\( z_2 = 2 + i \)[/tex], and [tex]\( z_3 = 3 - i \)[/tex]. Calculate:
#### (a) [tex]\( z_1 z_2 \)[/tex]
To find the product of [tex]\( z_1 \)[/tex] and [tex]\( z_2 \)[/tex]:
[tex]\[ z_1 z_2 = (1+i)(2+i) \][/tex]
Expanding the product:
[tex]\[ (1+i)(2+i) = 1 \cdot 2 + 1 \cdot i + i \cdot 2 + i \cdot i \][/tex]
[tex]\[ = 2 + i + 2i + i^2 \][/tex]
[tex]\[ = 2 + 3i + (-1) \][/tex]
[tex]\[ = 1 + 3i \][/tex]
So, [tex]\( z_1 z_2 = 1 + 3i \)[/tex].
#### (b) [tex]\( |z_3| \)[/tex]
To find the magnitude of [tex]\( z_3 \)[/tex]:
[tex]\[ z_3 = 3 - i \][/tex]
The magnitude is given by:
[tex]\[ |z_3| = \sqrt{(\text{Re}(z_3))^2 + (\text{Im}(z_3))^2} \][/tex]
[tex]\[ = \sqrt{3^2 + (-1)^2} \][/tex]
[tex]\[ = \sqrt{9 + 1} \][/tex]
[tex]\[ = \sqrt{10} \][/tex]
[tex]\[ \approx 3.162 \][/tex]
So, [tex]\( |z_3| \approx 3.162 \)[/tex].
#### (c) [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) \)[/tex]
To find the imaginary part of [tex]\( \frac{z_1}{z_2} \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ z_2 = 2 + i \][/tex]
[tex]\[ \frac{z_1}{z_2} = \frac{1 + i}{2 + i} \][/tex]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{1 + i}{2 + i} \cdot \frac{2 - i}{2 - i} \][/tex]
[tex]\[ = \frac{(1+i)(2-i)}{(2+i)(2-i)} \][/tex]
[tex]\[ = \frac{2 + i - 2i - i^2}{4 + 1} \][/tex]
[tex]\[ = \frac{2 - i - (-1)}{5} \][/tex]
[tex]\[ = \frac{3 - i}{5} \][/tex]
[tex]\[ = \frac{3}{5} - \frac{i}{5} \][/tex]
[tex]\[ = 0.6 - 0.2i \][/tex]
The imaginary part is:
[tex]\[ \operatorname{Im}\left(\frac{z_1}{z_2}\right) = -0.2 \][/tex]
So, [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) = 0.2 \)[/tex].
#### (d) [tex]\( \operatorname{Arg}(\bar{z}_1) \)[/tex]
To find the argument of the conjugate of [tex]\( z_1 \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ \bar{z}_1 = 1 - i \][/tex]
The argument is the angle the complex number makes with the positive real axis.
For [tex]\( 1 - i \)[/tex], it is in the fourth quadrant:
[tex]\[ \theta = \arctan\left(\frac{-1}{1}\right) \][/tex]
[tex]\[ = \arctan(-1) \][/tex]
[tex]\[ = -\frac{\pi}{4} \][/tex]
So, [tex]\( \operatorname{Arg}(\bar{z}_1) = -\frac{\pi}{4} \approx -0.785 \)[/tex].
#### (e) [tex]\( z_1 + 2 z_3 \)[/tex]
To find [tex]\( z_1 + 2z_3 \)[/tex]:
[tex]\[ z_1 = 1 + i \][/tex]
[tex]\[ z_3 = 3 - i \][/tex]
[tex]\[ z_1 + 2z_3 = (1 + i) + 2(3 - i) \][/tex]
[tex]\[ = 1 + i + 6 - 2i \][/tex]
[tex]\[ = 7 - i \][/tex]
So, [tex]\( z_1 + 2z_3 = 7 - i \)[/tex].
Hence:
1. [tex]\[ \text{Solutions to } z^3 - 2z - 4 = 0: \{2, -1 - i, -1 + i\} \][/tex]
2. [tex]\( z_1 z_2 = 1 + 3i \)[/tex]
3. [tex]\( |z_3| \approx 3.162 \)[/tex]
4. [tex]\( \operatorname{Im}\left(\frac{z_1}{z_2}\right) = 0.2 \)[/tex]
5. [tex]\( \operatorname{Arg}(\bar{z}_1) = -\frac{\pi}{4} \approx -0.785 \)[/tex]
6. [tex]\( z_1 + 2z_3 = 7 - 1i \)[/tex]
