2.1 Solve for \(x\) and \(y\) using complex numbers if given:

[tex]\[
\left|\begin{array}{cc}
xj & yj \\
-3 & -2
\end{array}\right| = \left|\begin{array}{cc}
x-y & j \\
-4 & 2
\end{array}\right|
\][/tex]

2.2 Given: \(z_1 = 4\angle 30^\circ\) and \(z_2 = 2\angle 60^\circ\), determine:

2.2.1 \(z_1 \cdot z_2\)

2.2.2 [tex]\(\frac{z_1}{z_2}\)[/tex], and leave the answer in polar form.



Answer :

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]