Certainly! Let's address each part of the question step-by-step:
### 2.1 Solve for \( x \) and \( y \) using complex numbers:
The given equation is:
[tex]\[ \left|\begin{array}{cc} x j & y j \\ -3 & -2 \end{array}\right| = \left|\begin{array}{cc} x - y & j \\ -4 & 2 \end{array}\right| \][/tex]
We need to evaluate the determinants of both matrices and then equate them.
Left-hand side (LHS):
[tex]\[ \left|\begin{array}{cc} x j & y j \\ -3 & -2 \end{array}\right| \][/tex]
To find the determinant:
[tex]\[ = (x j)(-2) - (y j)(-3) \][/tex]
[tex]\[ = -2x j + 3y j \][/tex]
[tex]\[ = j(-2x + 3y) \][/tex]
Right-hand side (RHS):
[tex]\[ \left|\begin{array}{cc} x - y & j \\ -4 & 2 \end{array}\right| \][/tex]
To find the determinant:
[tex]\[ = (x - y)(2) - (j)(-4) \][/tex]
[tex]\[ = 2(x - y) + 4j \][/tex]
Equate the determinants:
[tex]\[ j(-2x + 3y) = 2(x - y) + 4j \][/tex]
Separate the real and imaginary parts:
For imaginary parts:
[tex]\[ -2x + 3y = 4 \][/tex]
For real parts:
[tex]\[ 2(x - y) = 0 \][/tex]
[tex]\[ x - y = 0 \][/tex]
[tex]\[ x = y \][/tex]
Now substitute \( x = y \) into the imaginary parts equation:
[tex]\[ -2x + 3x = 4 \][/tex]
[tex]\[ x = 4 \][/tex]
Since \( x = y \), we also have \( y = 4 \).
Thus, the solutions are:
[tex]\[ x = 4 \][/tex]
[tex]\[ y = 4 \][/tex]
### 2.2 Given \( z_1 = 4 \breve{30^{\circ}} \) and \( z_2 = 2 \breve{60^{\circ}} \), determine:
#### 2.2.1 \( z_1 \cdot z_2 \)
To multiply two complex numbers in polar form, we multiply their moduli and add their angles.
- Modulus of \(z_1 \cdot z_2 \):
[tex]\[ |z_1| \cdot |z_2| = 4 \cdot 2 = 8 \][/tex]
- Angle of \(z_1 \cdot z_2 \):
[tex]\[ \arg(z_1) + \arg(z_2) = 30^\circ + 60^\circ = 90^\circ \][/tex]
So:
[tex]\[ z_1 \cdot z_2 = 8 \breve{90^\circ} \][/tex]
#### 2.2.2 \(\frac{z_1}{z_2}\)
To divide two complex numbers in polar form, we divide their moduli and subtract their angles.
- Modulus of \(\frac{z_1}{z_2} \):
[tex]\[ \frac{|z_1|}{|z_2|} = \frac{4}{2} = 2 \][/tex]
- Angle of \(\frac{z_1}{z_2} \):
[tex]\[ \arg(z_1) - \arg(z_2) = 30^\circ - 60^\circ = -30^\circ \][/tex]
So:
[tex]\[ \frac{z_1}{z_2} = 2 \breve{-30^\circ} \][/tex]
In summary:
2.1 Solutions:
[tex]\[ x = 4 \][/tex]
[tex]\[ y = 4 \][/tex]
2.2 Solutions:
[tex]\[
\begin{aligned}
&2.2.1 \quad z_1 \cdot z_2 = 8 \breve{90^\circ} \\
&2.2.2 \quad \frac{z_1}{z_2} = 2 \breve{-30^\circ}
\end{aligned}
\][/tex]
