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Complex z and w with z + w = conj(z)/w and 1/w + z = w conj(z) | JEE

JEE Maths question with a full step-by-step solution.

Question
Consider complex numbers zz and ww satisfying the equations
z+w=zˉwand1w+z=wzˉ,z + w = \frac{\bar z}{w} \qquad \text{and} \qquad \frac{1}{w} + z = w\bar z ,
 then which of the following is/are correct?
AThe value of ∣z∣+∣w∣|z| + |w| is 22
BIf arg⁡(z)=π\arg(z) = \pi then arg⁡(w)=−π3\arg(w) = -\dfrac{\pi}{3} or π3\dfrac{\pi}{3}correct
CIf w=1w = 1, then Re⁡(z)=12\operatorname{Re}(z) = \dfrac12
DIf w=−1w = -1, then Re⁡(z)=12\operatorname{Re}(z) = \dfrac12correct
Solution
Step 1: Clear the denominators by multiplying each equation by ww (note w≠0w \ne 0).
zw+w2=zˉ...(i)zw + w^{2} = \bar z \qquad \text{...(i)}
1+zw=w2zˉ...(ii)1 + zw = w^{2}\bar z \qquad \text{...(ii)}
 Step 2: From (i), zw=zˉ−w2zw = \bar z - w^{2}. Substitute this into (ii).
1+zˉ−w2=w2zˉ.1 + \bar z - w^{2} = w^{2}\bar z .
 Step 3: Bring everything to one side and factor by grouping.
1+zˉ−w2−w2zˉ=0  ⟹  (1+zˉ)−w2(1+zˉ)=01 + \bar z - w^{2} - w^{2}\bar z = 0 \implies \left(1 + \bar z\right) - w^{2}\left(1 + \bar z\right) = 0
  ⟹  (1+zˉ)(1−w2)=0.\implies \left(1 + \bar z\right)\left(1 - w^{2}\right) = 0 .
 Step 4: So at least one factor vanishes - two cases. Step 5: Case I, zˉ=−1\bar z = -1, i.e. z=−1z = -1 (so arg⁡z=π\arg z = \pi). Put z=−1z = -1 into (i):
−w+w2=−1  ⟹  w2−w+1=0  ⟹  w=1±i32.-w + w^{2} = -1 \implies w^{2} - w + 1 = 0 \implies w = \frac{1 \pm i\sqrt3}{2} .
 These are cos⁡π3±isin⁡π3\cos\dfrac{\pi}{3} \pm i\sin\dfrac{\pi}{3}, so arg⁡(w)=±π3\arg(w) = \pm\dfrac{\pi}{3}. That is exactly option (2), so (2) is correct. Step 6: Case II, w2=1w^{2} = 1, i.e. w=1w = 1 or w=−1w = -1. Step 7: Try w=1w = 1 in (i): z+1=zˉz + 1 = \bar z. Writing z=x+iyz = x+iy,
(x+1)+iy=x−iy  ⟹  x+1=x,(x+1) + iy = x - iy \implies x + 1 = x ,
 which is impossible. So w=1w = 1 never occurs and (3) is not correct. Step 8: Try w=−1w = -1 in (i): −z+1=zˉ-z + 1 = \bar z, i.e.
1=z+zˉ=2Re⁡(z)  ⟹  Re⁡(z)=12.1 = z + \bar z = 2\operatorname{Re}(z) \implies \operatorname{Re}(z) = \frac12 .
 So (4) is correct. Step 9: Test (A). In Case I, ∣z∣=1|z| = 1 and ∣w∣=1|w| = 1, giving ∣z∣+∣w∣=2|z|+|w| = 2; but in Case II, w=−1w = -1 and z=12+iyz = \dfrac12 + iy for any real yy, so ∣z∣=14+y2|z| = \sqrt{\frac14 + y^{2}} is not fixed and ∣z∣+∣w∣|z|+|w| is not always 22. So (1) is not correct. Answer: (2) and (4).
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