Complex Numbers (JEE Main)

Question 1

If lvert zrvert =1, then the maximum value of lvert z−1rvert lvert z+1rvert ² is

Accepted Answer

C. Put z=cosθ+isinθ: lvert z−1rvert ²=2−2cosθ and lvert z+1rvert ²=2+2cosθ. Let u=1+cosθ∈[0,2]. Then (lvert z−1rvert lvert z+1rvert ²)²=2(2−u)·4u²=32·u/2·u/2·(2−u). By AM-GM, u/2·u/2·(2−u)≤(2/3)³=8/27, with equality at u=4/3 (i.e. cosθ=1/3). So the square is at most 256/27 and the maximum is 16/(3√3)=(16√3)/9. Trap: 8 maximises lvert z−1rvert ≤2 and lvert z+1rvert ²≤4 separately, but they peak at opp

Question 2

A non-zero complex number z satisfies lvert z−6/zrvert =1. If M and m are the greatest and least possible values of lvert zrvert, find M²+m².

Accepted Answer

13. Let r=lvert zrvert. By the triangle inequality, 1=lvert z−6/zrvert ≥lvert r−6/rrvert. So r−6/r≤1⇒ r²−r−6≤0⇒ r≤3, and 6/r−r≤1⇒ r²+r−6≥0⇒ r≥2. Both bounds are attained: z=3 gives 3−2=1; z=−2 gives −2+3=1. M=3, m=2, M²+m²=13. Trap: using only the upper bound, or forgetting to check that the bounds are actually attained. Exam pattern: JEE Main – maximum and minimum of modulus with a z and 1/z conditio

Question 3

The locus of z satisfying arg((z−3i)/(z+3))=π/4 is an arc of a circle. The maximum value of lvert zrvert on this locus is

Accepted Answer

C. With A=−3, B=3i, the condition says the angle at z subtended by AB is π/4, so z is on an arc of a circle of radius (AB)/(2sin(π/4))=(3√2)/(√2)=3. The two possible centres are 0 and −3+3i. Test: z=−6+3i lies on lvert z+3−3irvert =3 and (z−3i)/(z+3)=(−6)/(−3+3i)=1+i, whose argument is π/4. On lvert zrvert =3, e.g. z=3 gives arg(3−3i)/6=−π/4, so that circle is wrong. So the locus is the major arc of

Complex Numbers: questions with solutions

Get the full chapter for ₹99