Question

If z and w are two complex numbers such that $|zw| = 1$ and $\arg(z) – \arg(w) = \frac{\pi}{2}$, then :

• A. $z\overline{w} = \frac{1-i}{\sqrt{2}}$
• B. $z\overline{w} = i$
• C. $z\overline{w} = \frac{-1+i}{\sqrt{2}}$
• D. $z\overline{w} = -i$

Answer 1

Answer: (B)

$z\overline{w} = i$

Answer 2

Given $|zw|=1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$. Let $z = |z|e^{i\theta_1}$ and $w = |w|e^{i\theta_2}$, where $\theta_1 = \arg(z)$ and $\theta_2 = \arg(w)$. From $|zw|=1$, we have $|z||w|=1$. The condition on arguments is $\theta_1 - \theta_2 = \frac{\pi}{2}$.

We need to find $z\overline{w}$. The conjugate of $w$ is $\overline{w} = |w|e^{-i\theta_2}$.

Now, we compute the product $z\overline{w}$: $z\overline{w} = (|z|e^{i\theta_1}) \times (|w|e^{-i\theta_2})$ $z\overline{w} = |z||w| \times e^{i\theta_1}e^{-i\theta_2}$ $z\overline{w} = |z||w| e^{i(\theta_1 - \theta_2)}$

Substitute the given values: $|z||w|=1$ and $\theta_1 - \theta_2 = \frac{\pi}{2}$. $z\overline{w} = 1 \times e^{i\frac{\pi}{2}}$ $z\overline{w} = e^{i\frac{\pi}{2}}$

Using Euler's formula, $e^{i\phi} = \cos\phi + i\sin\phi$: $z\overline{w} = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)$ $z\overline{w} = 0 + i(1)$ $z\overline{w} = i$