Complex Numbers (JEE Advanced)

Question 1

Let θ∈(π,2π) and z=1+cosθ+isinθ. Here Arg denotes the principal argument, with values in (−π,π]. The set of all θ∈(π,2π) for which Arg(z³)=(3θ)/2−π is ______. Exactly one option is correct.

Accepted Answer

B. z=1+e^(iθ)=e^(iθ/2)(e^(−iθ/2)+e^(iθ/2))=2cosθ/2 e^(iθ/2).
For θ∈(π,2π), cosθ/2<0, so z=2lvertcosθ/2rvert e^(i(θ/2−π)) with θ/2−π∈(−π/2,0); thus Argz=θ/2−π.
Then z³ has the argument value v=(3θ)/2−3π∈(−(3π)/2,0), which may lie outside (−π,π].
If v>−π, i.e. θ>(4π)/3: Arg(z³)=v=(3θ)/2−3π≠(3θ)/2−π.
If v≤−π, i.e. θ≤(4π)/3: Arg(z³)=v+2π=(3θ)/2−π.
At θ=(4π)/3, v=−π is not a principal value; the principal

Question 2

Let S={z∈mathbbCsetminus{1,−1}: Arg(z−1)+Arg(z+1)=π}. Here Arg denotes the principal argument, with values in (−π,π]. Then S equals ______. Exactly one option is correct.

Accepted Answer

D. Each principal argument lies in (−π,π]. If the sum equals π, then π is an argument of (z−1)(z+1)=z²−1, so z²−1 is a negative real number: z² is real and z²<1.
So either z=x is real with −1<x<1, or z=iy with y≠0 (then z²=−y²<0<1). These are only candidates; each must be checked, because the sum of principal values can differ from Arg(z²−1) by 2π.
Real x∈(−1,1): x−1<0, x+1>0, so the sum is π+0=π. In

Question 3

The complex number z ranges over the closed disc lvert z−2rvert≤1, and u=z²−4z+7. Here Arg denotes the principal argument, with values in (−π,π]. The greatest value of Arg(u) is ______. Exactly one op

Accepted Answer

A. u=(z−2)²+3. Write z−2=re^(iφ) with 0≤ r≤1; then (z−2)²=r²e^(2iφ).
As r runs over [0,1] and φ over [0,2π), the value r²e^(2iφ) runs over every complex number of modulus at most 1. So u runs over the whole closed disc lvert u−3rvert≤1, which does not contain 0.
The largest argument of a point of this disc is reached at the point of contact of the upper tangent from the origin: sinα=1/3, so tanα=1/(√

Complex Numbers: questions with solutions

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