سوال ۲۶ کنکور ارشد علوم کامپیوتر ۱۴۰۳

The equation contains $\bar z$ and $\mathrm{Im}(z)$, so it is not a polynomial in $z$ — split it into real and imaginary parts. Put $z = x + iy$:

• $z^2 = (x^2 - y^2) + 2xyi$
• $4\bar z = 4x - 4yi$
• $2z = 2x + 2yi$
• $i\,\mathrm{Im}(z) = iy$

Imaginary part $= 0$: $\;2xy - 4y + 2y + y = y(2x - 1) = 0 \Rightarrow y = 0 \text{ or } x = \tfrac12.$

Real part $= 1$: $\;x^2 - y^2 + 6x = 1.$

• If $y = 0$: $x^2 + 6x - 1 = 0 \Rightarrow x = -3 \pm \sqrt{10}$ — 2 roots.
• If $x = \tfrac12$: $\tfrac14 - y^2 + 3 = 1 \Rightarrow y^2 = \tfrac94 \Rightarrow y = \pm\tfrac32$ — 2 roots.

In total there are $\boxed{4}$ roots, so the answer is option 4.

Pitfall: the presence of $\bar z$ / $\mathrm{Im}(z)$ breaks analyticity — counting via "degree of a polynomial" is wrong. Always convert to $x,y$.