Question:
If $z = \frac{2 - i}{i}$, then $\text{Re}(z^2) + \text{Im}(z^2)$ is equal to:
If $z = \frac{2 - i}{i}$, then $\text{Re}(z^2) + \text{Im}(z^2)$ is equal to:
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Always remember $\frac{1}{i} = -i$. It simplifies complex fractions instantly without needing conjugates.
Updated On: May 2, 2026
- $1$
- $-1$
- $2$
- $-2$
- $3$
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The Correct Option is D
Solution and Explanation
Concept:
To simplify a complex fraction, use the identity $\frac{1}{i} = -i$. Then square the complex number and extract real and imaginary parts.
Step 1: Simplify $z$.
\[ z = \frac{2 - i}{i} = (2 - i)\left(\frac{1}{i}\right) = (2 - i)(-i) \] \[ z = -2i + i^2 = -2i - 1 = -1 - 2i \]
Step 2: Calculate $z^2$.
\[ z^2 = (-1 - 2i)^2 \] \[ = (-1)^2 + 2(-1)(-2i) + (-2i)^2 \] \[ = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i \]
Step 3: Find the required sum.
\[ \text{Re}(z^2) = -3, \quad \text{Im}(z^2) = 4 \] \[ \text{Re}(z^2) + \text{Im}(z^2) = -3 + 4 = 1 \]
Step 1: Simplify $z$.
\[ z = \frac{2 - i}{i} = (2 - i)\left(\frac{1}{i}\right) = (2 - i)(-i) \] \[ z = -2i + i^2 = -2i - 1 = -1 - 2i \]
Step 2: Calculate $z^2$.
\[ z^2 = (-1 - 2i)^2 \] \[ = (-1)^2 + 2(-1)(-2i) + (-2i)^2 \] \[ = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i \]
Step 3: Find the required sum.
\[ \text{Re}(z^2) = -3, \quad \text{Im}(z^2) = 4 \] \[ \text{Re}(z^2) + \text{Im}(z^2) = -3 + 4 = 1 \]
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