Question

If z and w are two nonzero complex numbers such that |zw| = 1 and arg z−arg w = $\dfrac{\pi }{2}$,​ then $\overline{z}w$ is equal to
(a) −1
(b) i
(c) −i
(d) 1

Answer 1

Hint: To solve this problem, we should know the basics of complex numbers. We should also know a few properties of complex number; these are –
(1) |zw| = |z||w| (that is the modulus of the product of two complex number is equal to product of modulus of each complex number individual)
(2) cis A = cosA + isinA
(3) For a complex number z with z = ∣z∣ (cosA + isinA), arg(z) = A.
(4) $cis\left( -\dfrac{\pi }{2} \right)$ = -i

Complete step-by-step answer:

To solve the problem, we start from the given conditions, we have,
|zw| = 1 and arg(z)−arg(w) = ​$\dfrac{\pi }{2}$

Now, we let |z| = r.
Now, we use the property |zw| = |z||w|, thus, we have,
Since, |zw| = 1, |w| = $\dfrac{1}{|z|}$ = $\dfrac{1}{r}$
Now, we let z = ∣z∣(cosA + isinA) = r cisA. (Using, cis A = cosA + isinA). Also, we have -
w = ∣w∣(cosB + isinB) = ​$\dfrac{1}{r}$ cisB
From the problem in hand, we have arg(z)−arg(w) = ​$\dfrac{\pi }{2}$.
Using the property that –
For a complex number z with z = ∣z∣ (cosA + isinA), arg(z) = A. Thus, in this case, we have,
arg(z) = A and arg(w) = B, thus, we have
A−B = $\dfrac{\pi }{2}$
$\overline{z}w$ = (rcis(−A)) $\times$ $\dfrac{1}{r}$ cisB = cis(B-A) = cis(-(A-B))
$\overline{z}w$ = $cis\left( -\dfrac{\pi }{2} \right)$ = -i
Hence, the correct answer is (c) –i.

Note: While solving a problem related to complex numbers, we should be aware about the formula of complex numbers which would be useful to solve the respective problem. For example, in some cases, we represent z = r (cosA + isinA). In other cases, it is useful to use the Euler’s representation for complex numbers. In Euler’s representation, we represent z = $r{{e}^{iA}}$ (which is the same as r (cosA + isinA)). Although, both can be used to solve the problem, depending on the problem, one of them would make the task less tedious.