Complex Numbers
Algebra of Complex Numbers
Grade Class 11

Question:

<p>Let \(|z|=2\) and \(\text{Re}(z^2)=0\). Then \(z\) equals:</p>
2e^(i\pi/4)
\sqrt{2} \cdot e^(i\pi/4)
2e^(i\pi/6)
i\sqrt{2}

Step-by-Step Solution

Key Concept: Re(z^2)=0 with |z|=2: let z=2e^(i\theta), z^2=4e^(2i\theta), Re=4cos2\theta=0 \Rightarrow 2\theta=\pi/2+k\pi \Rightarrow \theta=\pi/4, 3\pi/4, 5\pi/4, 7\pi/4. So z=2e^(i\pi/4) is one valid solution.
<p>$z=2e^{i\theta}$, $z^2=4e^{2i\theta}$. $\text{Re}(z^2)=4\cos2\theta=0\Rightarrow\theta=\pi/4+k\pi/2$. For $\theta=\pi/4$: $z=2e^{i\pi/4}$. ✓ Answer A.</p>
Correct Answer: A

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