Given: $$ \begin{array}{l} z_{1}=3-2 i \ z_{2}=4+3 i \ z_{3}=a+b i \end{array} $$ 1.3.1 Prove that $$ \frac{z_{1} \cdot z_{3}}{z_{2}}=\frac{6 a+17 b}{25}+\frac{-17 a+6 b}{25} i $$ 1.3.2 Given: $$ \frac{z_{1} \cdot z_{3}}{z_{2}}=\frac{28}{25}+\frac{29}{25} i $$ Solve $$ a $$ and $$ b $$.

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### **1.3.1 Prove that**
$$
\frac{z_{1} \cdot z_{3}}{z_{2}} = \frac{6a + 17b}{25} + \frac{-17a + 6b}{25}i
$$
**Given:**
$$
\begin{cases}
z_{1} = 3 - 2i \
z_{2} = 4 + 3i \
z_{3} = a + bi
\end{cases}
$$

**Proof:**

1. **Compute $$ z_{1} \cdot z_{3} $$:**
   $$
   z_{1} \cdot z_{3} = (3 - 2i)(a + bi)
   $$
   Apply the distributive property:
   $$
   = 3a + 3bi - 2a i - 2b i^2
   $$
   Since $$ i^2 = -1 $$:
   $$
   = 3a + 3bi - 2a i + 2b
   $$
   Combine like terms:
   $$
   = (3a + 2b) + (3b - 2a)i
   $$

2. **Divide by $$ z_{2} $$:**
   $$
   \frac{z_{1} \cdot z_{3}}{z_{2}} = \frac{(3a + 2b) + (3b - 2a)i}{4 + 3i}
   $$
   To simplify, multiply the numerator and denominator by the conjugate of the denominator $$ 4 - 3i $$:
   $$
   = \frac{[(3a + 2b) + (3b - 2a)i](4 - 3i)}{(4 + 3i)(4 - 3i)}
   $$
   
3. **Simplify the denominator:**
   $$
   (4 + 3i)(4 - 3i) = 4^2 - (3i)^2 = 16 - 9(-1) = 16 + 9 = 25
   $$
   
4. **Expand the numerator:**
   $$
   [(3a + 2b) + (3b - 2a)i](4 - 3i) = (3a + 2b)(4) + (3a + 2b)(-3i) + (3b - 2a)i(4) + (3b - 2a)i(-3i)
   $$
   $$
   = 12a + 8b - 9a i - 6b i + 12b i - 8a i - 9b i^2 + 6a i^2
   $$
   Again, using $$ i^2 = -1 $$:
   $$
   = 12a + 8b - 9a i - 6b i + 12b i - 8a i + 9b - 6a
   $$
   Combine like terms:
   $$
   = (12a - 6a) + (8b + 9b) + (-9a i - 8a i) + (-6b i + 12b i)
   $$
   $$
   = 6a + 17b - 17a i + 6b i
   $$
   $$
   = (6a + 17b) + (-17a + 6b)i
   $$
   
5. **Combine the results:**
   $$
   \frac{z_{1} \cdot z_{3}}{z_{2}} = \frac{6a + 17b}{25} + \frac{-17a + 6b}{25}i
   $$
   
**Thus, the proof is complete.**

---

### **1.3.2 Solve for $$ a $$ and $$ b $$**

**Given:**
$$
\frac{z_{1} \cdot z_{3}}{z_{2}} = \frac{28}{25} + \frac{29}{25}i
$$
From part 1.3.1, we have:
$$
\frac{z_{1} \cdot z_{3}}{z_{2}} = \frac{6a + 17b}{25} + \frac{-17a + 6b}{25}i
$$
Set the real and imaginary parts equal to each other:
$$
\begin{cases}
\frac{6a + 17b}{25} = \frac{28}{25} \
\frac{-17a + 6b}{25} = \frac{29}{25}
\end{cases}
$$
Multiplying both equations by 25 to eliminate denominators:
$$
\begin{cases}
6a + 17b = 28 \
-17a + 6b = 29
\end{cases}
$$

**Solve the system of equations:**

1. **Write the system:**
   $$
   \begin{cases}
   6a + 17b = 28 \quad \text{(1)} \
   -17a + 6b = 29 \quad \text{(2)}
   \end{cases}
   $$

2. **Eliminate one variable:**

To eliminate $$ a $$, multiply equation (1) by 17 and equation (2) by 6:
   $$
   \begin{cases}
   17 \times (6a + 17b) = 17 \times 28 \
   6 \times (-17a + 6b) = 6 \times 29
   \end{cases}
   $$
   $$
   \begin{cases}
   102a + 289b = 476 \quad \text{(3)} \
   -102a + 36b = 174 \quad \text{(4)}
   \end{cases}
   $$
   
3. **Add equations (3) and (4) to eliminate $$ a $$:**
   $$
   102a + 289b - 102a + 36b = 476 + 174
   $$
   $$
   325b = 650
   $$
   $$
   b = \frac{650}{325} = 2
   $$

4. **Substitute $$ b = 2 $$ into equation (1):**
   $$
   6a + 17(2) = 28
   $$
   $$
   6a + 34 = 28
   $$
   $$
   6a = 28 - 34 = -6
   $$
   $$
   a = \frac{-6}{6} = -1
   $$

**Solution:**
$$
a = -1 \quad \text{and} \quad b = 2
$$

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**Therefore, the values are $$ a = -1 $$ and $$ b = 2 $$.**
$$ a = -1 $$ and $$ b = 2 $$.

Given: \[ \begin{array}{l} z_{1}=3-2 i \\ z_{2}=4+3 i \\ z_{3}=a+b i \end{array} \] 1.3.1 Prove that \( \frac{z_{1} \cdot z_{3}}{z_{2}}=\frac{6 a+17 b}{25}+\frac{-17 a+6 b}{25} i \) 1.3.2 Given: \[ \frac{z_{1} \cdot z_{3}}{z_{2}}=\frac{28}{25}+\frac{29}{25} i \] Solve \( a \) and \( b \).

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