If z=2-3i, then show that `z^(2)-4z+13=0`.

Answer

To show that \( z^2 - 4z + 13 = 0 \) for \( z = 2 - 3i \), we will substitute the value of \( z \) into the expression and simplify it step by step.

### Step 1: Substitute \( z \) into the expression
We start with the expression:
\[ z^2 - 4z + 13 \]
Substituting \( z = 2 - 3i \):
\[ (2 - 3i)^2 - 4(2 - 3i) + 13 \]

### Step 2: Calculate \( z^2 \)
Now we need to calculate \( (2 - 3i)^2 \):
\[ (2 - 3i)^2 = 2^2 - 2 \cdot 2 \cdot 3i + (3i)^2 \]
Calculating each term:
- \( 2^2 = 4 \)
- \( -2 \cdot 2 \cdot 3i = -12i \)
- \( (3i)^2 = 9i^2 = 9(-1) = -9 \)

Putting it all together:
\[ (2 - 3i)^2 = 4 - 12i - 9 = -5 - 12i \]

### Step 3: Calculate \( -4z \)
Next, we calculate \( -4(2 - 3i) \):
\[ -4(2 - 3i) = -8 + 12i \]

### Step 4: Combine the results
Now we substitute back into the expression:
\[ (-5 - 12i) + (-8 + 12i) + 13 \]

Combining like terms:
- Real parts: \( -5 - 8 + 13 = 0 \)
- Imaginary parts: \( -12i + 12i = 0 \)

Thus, we have:
\[ 0 + 0i = 0 \]

### Conclusion
We have shown that:
\[ z^2 - 4z + 13 = 0 \]
This confirms that the equation holds true for \( z = 2 - 3i \).