Question:medium

If the conjugate of a complex number $z$ is $\frac{1}{z - i}$, then $z$ can be:

Show Hint

Since $|z|^2 - i\bar{z} = 1$ is real, $i\bar{z}$ must be real. This immediately implies that $\bar{z}$ (and thus $z$) has no real part and is purely imaginary, which eliminates options (C) and (D).
Updated On: Jun 3, 2026
  • $i\left(\frac{1+\sqrt{5}}{2}\right)$
  • $i\left(\frac{1-\sqrt{5}}{2}\right)$
  • $\frac{1+i\sqrt{5}}{2}$
  • $\frac{1-i\sqrt{5}}{2}$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use that $z\bar z$ is real.
For any complex number, $z\bar z = |z|^2$ is a real number. We will use this to pin down the form of $z$.

Step 2: Clear the fraction.
We are given $\bar z = \frac{1}{z - i}$. Multiply both sides by $(z-i)$: \[ \bar z (z - i) = 1 \implies |z|^2 - i\bar z = 1 \]

Step 3: Isolate the $i\bar z$ term.
Rearrange: \[ i\bar z = |z|^2 - 1 \] The right side is real, so the left side $i\bar z$ must be real too.

Step 4: Force $z$ to be imaginary.
Write $z = x + iy$, so $i\bar z = i(x - iy) = y + ix$. For this to be real we need $x = 0$. So $z = iy$.

Step 5: Plug back in.
With $z = iy$, the relation $\bar z(z-i)=1$ gives $-iy\,(iy - i) = 1$, which simplifies to \[ y(y-1) = 1 \implies y^2 - y - 1 = 0 \]

Step 6: Solve for $y$.
By the quadratic formula $y = \frac{1 \pm \sqrt5}{2}$. So one allowed value is \[ \boxed{ z = i\left(\frac{1+\sqrt5}{2}\right) } \]
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