Skip to content
Exercise 2.9 · Q12

Q.If zz is a complex number such that z∈C∖Rz\in\mathbb C\setminus\mathbb R and z+1z∈Rz+\dfrac1z\in\mathbb R, then ∣z∣|z| is

(1) 00
(2) 11
(3) 22
(4) 33
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
50% · 61/122 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

z+1zz+\dfrac1z being real means its imaginary part is zero; computing that imaginary part in terms of x,yx,y directly and factoring out yy (which is nonzero because zz is not real) forces ∣z∣=1|z|=1.

Step 1. Write z=x+iyz=x+iy with y≠0y\ne0 (since z∈C∖Rz\in\mathbb C\setminus\mathbb R means zz is not real).

Step 2. Compute 1z\dfrac1z in standard form.

1z=1x+iy=x−iyx2+y2.\frac1z=\frac1{x+iy}=\frac{x-iy}{x^2+y^2}.

Step 3. Form z+1zz+\dfrac1z and isolate the imaginary part.

z+1z=(x+xx2+y2)+i(y−yx2+y2).z+\frac1z=\left(x+\frac{x}{x^2+y^2}\right)+i\left(y-\frac{y}{x^2+y^2}\right).

So Im⁡(z+1z)=y−yx2+y2=y(1−1x2+y2)\operatorname{Im}\left(z+\dfrac1z\right)=y-\dfrac{y}{x^2+y^2}=y\left(1-\dfrac1{x^2+y^2}\right). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.