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प्रश्न
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = `pi/2`, then show that `barz`w = –i.
उत्तर
Let z = r1 (cosθ1 + isinθ1) and w = r2 (cosθ2 + isinθ2)
zw = r1r2 [(cosθ1 + isinθ1)] [(cosθ2 + isinθ2)]
|zw| = r1r2 = 1
Now arg(z) – arg(w) = `pi/2`
θ1 – θ2 = `pi/2`
⇒ arg `(z/w) = pi/2`
`barzw` = r1 (cosθ1 – isinθ1) r2 (cosθ2 + isinθ2)
= r1 r2 [cosθ1 cosθ2 + icosθ1 sinθ2 – isinθ1 cosθ2 – i2 sinθ1 sinθ2]
= r1 r2 [(cosθ1 cosθ2 + sinθ1 sinθ2) + i(cosθ1 sinθ2 – sinθ1 cosθ2)]
= r1 r2 [cos(θ2 – θ1) + isin(θ2 – θ1)]
= `r_1r_2 [cos((-pi)/2) + i sin((-pi)/2)]`
= `r_1r_2 [cos pi/2 - i sin pi/2]`
= 1 .....[0 – i]
Here `barzw` = –i.
Hence proved.
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