Find Re `((z_(1)z_(2))/(z_(1))),` give `z_(1)=2-i` and `z_(2)=-2+i`

Answer

To solve the problem of finding the real part of \(\frac{z_1 z_2}{z_1}\) given \(z_1 = 2 - i\) and \(z_2 = -2 + i\), we will follow these steps:

### Step 1: Calculate \(z_1 z_2\)

Given:
- \(z_1 = 2 - i\)
- \(z_2 = -2 + i\)

We will multiply \(z_1\) and \(z_2\):

\[ z_1 z_2 = (2 - i)(-2 + i) \]

Using the distributive property (FOIL method):

\[ = 2 \cdot (-2) + 2 \cdot i - i \cdot (-2) - i \cdot i \]
\[ = -4 + 2i + 2i - i^2 \]
\[ = -4 + 4i - (-1) \quad (\text{since } i^2 = -1) \]
\[ = -4 + 4i + 1 \]
\[ = -3 + 4i \]

### Step 2: Divide \(z_1 z_2\) by \(z_1\)

Now we need to compute:

\[ \frac{z_1 z_2}{z_1} = \frac{-3 + 4i}{2 - i} \]

### Step 3: Rationalize the denominator

To simplify this expression, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{-3 + 4i}{2 - i} \cdot \frac{2 + i}{2 + i} \]

Calculating the denominator:

\[ (2 - i)(2 + i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]

Calculating the numerator:

\[ (-3 + 4i)(2 + i) = -3 \cdot 2 - 3 \cdot i + 4i \cdot 2 + 4i \cdot i \]
\[ = -6 - 3i + 8i + 4i^2 \]
\[ = -6 + 5i + 4(-1) \quad (\text{since } i^2 = -1) \]
\[ = -6 + 5i - 4 \]
\[ = -10 + 5i \]

Thus, we have:

\[ \frac{-3 + 4i}{2 - i} = \frac{-10 + 5i}{5} = -2 + i \]

### Step 4: Find the real part

The real part of \(-2 + i\) is:

\[ \text{Re}\left(-2 + i\right) = -2 \]

### Final Answer

The real part of \(\frac{z_1 z_2}{z_1}\) is \(-2\).

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