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06/01/2022 12:34 pm
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If \(\frac{z+i}{z+2i}\) is purely real, then the locus of z is
(a) x - axis
(b) y - axis
(c) y = x
(d) y = \(\frac{x}{2}\)
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06/01/2022 12:41 pm
Correct answer: (b) y - axis
Explanation:
\(\frac{z+i}{z+2i}\) is purely real, so \(\frac{z+i}{z+2i}\) = \(\Big(\overline{\frac{z+i}{z+2i}}\Big)\)
⇒ \(z \bar z\) - 2iz + \(i \bar z\) + 2 = \(z \bar z\) - iz + \(2i \bar z\) + 2
⇒ \(z(-i) + \bar z(-i)\) = 0
⇒ z = \(- \bar z\)
z is purely imaginary
Therefore, z lies on y - axis.