Sure, let's solve each part step-by-step.
### 2.1 Simplify [tex]\(3 j^{12} + 2 j^{23}\)[/tex] to rectangular form
First, recall that [tex]\( j \)[/tex] is the imaginary unit, defined as [tex]\( j = \sqrt{-1} \)[/tex].
Here's a useful cycle for powers of [tex]\( j \)[/tex]:
- [tex]\( j^1 = j \)[/tex]
- [tex]\( j^2 = -1 \)[/tex]
- [tex]\( j^3 = -j \)[/tex]
- [tex]\( j^4 = 1 \)[/tex]
- This cycle repeats every 4, so [tex]\( j^n = j^{(n \mod 4)} \)[/tex].
Now let's simplify:
- [tex]\( j^{12} \)[/tex]: [tex]\( 12 \mod 4 = 0 \)[/tex] (remainder is 0), so [tex]\( j^{12} = j^0 = 1 \)[/tex].
- [tex]\( j^{23} \)[/tex]: [tex]\( 23 \mod 4 = 3 \)[/tex] (remainder is 3), so [tex]\( j^{23} = j^3 = -j \)[/tex].
Substitute back into the expression:
[tex]\[ 3 j^{12} + 2 j^{23} = 3 \cdot 1 + 2 \cdot (-j) = 3 - 2j \][/tex]
So, the rectangular form of [tex]\( 3 j^{12} + 2 j^{23} \)[/tex] is:
[tex]\[ \boxed{3 - 2j} \][/tex]
### 2.2 Solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] if [tex]\( \frac{a}{b} + j(a - b) = 7 + 2j \)[/tex]
Equate the real and imaginary parts on both sides of the equation:
[tex]\[ \text{Real part: } \frac{a}{b} = 7 \][/tex]
[tex]\[ \text{Imaginary part: } a - b = 2 \][/tex]
From the real part:
[tex]\[ \frac{a}{b} = 7 \implies a = 7b \][/tex]
Substitute [tex]\(a = 7b\)[/tex] into the imaginary part:
[tex]\[ 7b - b = 2 \implies 6b = 2 \implies b = \frac{2}{6} = \frac{1}{3} \][/tex]
Now substitute [tex]\(b = \frac{1}{3}\)[/tex] back to find [tex]\(a\)[/tex]:
[tex]\[ a = 7b = 7 \times \frac{1}{3} = \frac{7}{3} \][/tex]
So, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ \boxed{a = \frac{7}{3}, \quad b = \frac{1}{3}} \][/tex]
### 2.3 Convert [tex]\(12 - 5j\)[/tex] to polar form
To convert a complex number [tex]\(z = a + bj\)[/tex] to polar form, use the formula:
[tex]\[ r = \sqrt{a^2 + b^2} \][/tex]
[tex]\[ \theta = \arctan\left(\frac{b}{a}\right) \text{ (in radians)} \][/tex]
For [tex]\(z = 12 - 5j\)[/tex]:
- [tex]\(a = 12\)[/tex]
- [tex]\(b = -5\)[/tex]
Calculate [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \][/tex]
Calculate [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{-5}{12}\right) \][/tex]
[tex]\[ \theta \approx \arctan\left(-0.4167\right) \][/tex]
[tex]\[ \theta \approx -0.3927 \text{ radians} \][/tex]
Since we want [tex]\( \theta \)[/tex] to be positive:
[tex]\[ \theta = 2\pi - 0.3927 \approx 5.8905 \text{ radians} \][/tex]
So the polar form of [tex]\(12 - 5j\)[/tex] is:
[tex]\[ \boxed{(13, 5.8905)} \][/tex]
### 2.4 Solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] using Cramer's rule
Given the system of linear equations:
[tex]\[ 5x + 2y = -19 \][/tex]
[tex]\[ 3x + 4y = -17 \][/tex]
Cramer's rule states that for a system [tex]\( A \mathbf{x} = \mathbf{B} \)[/tex]:
[tex]\[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)} \][/tex]
Where [tex]\( A \)[/tex] is the coefficient matrix:
[tex]\[ A = \begin{pmatrix}
5 & 2 \\
3 & 4
\end{pmatrix} \][/tex]
The determinants are calculated as follows:
1. [tex]\( \det(A) = (5 \cdot 4) - (2 \cdot 3) = 20 - 6 = 14 \)[/tex]
2. [tex]\( \det(A_x) \)[/tex] replaces the first column of [tex]\( A \)[/tex] with vector [tex]\( B \)[/tex]:
[tex]\[ A_x = \begin{pmatrix}
-19 & 2 \\
-17 & 4
\end{pmatrix} \][/tex]
[tex]\[ \det(A_x) = (-19 \cdot 4) - (2 \cdot -17) = -76 + 34 = -42 \][/tex]
3. [tex]\( \det(A_y) \)[/tex] replaces the second column of [tex]\( A \)[/tex] with vector [tex]\( B \)[/tex]:
[tex]\[ A_y = \begin{pmatrix}
5 & -19 \\
3 & -17
\end{pmatrix} \][/tex]
[tex]\[ \det(A_y) = (5 \cdot -17) - (3 \cdot -19) = -85 + 57 = -28 \][/tex]
Now compute [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{\det(A_x)}{\det(A)} = \frac{-42}{14} = -3 \][/tex]
[tex]\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-28}{14} = -2 \][/tex]
So, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are:
[tex]\[ \boxed{x = -3, \quad y = -2} \][/tex]
