JEE Challenger
More from Complex Numbers and Quadratic Equations
MathHardAdv 2019 • Paper 1

Principal Argument of Complex Expression with Maximum Modulus Condition

Consider SS to be the set containing all complex numbers zz that satisfy ∣z−2+i∣≥5|z - 2 + i| \ge \sqrt{5}. If z0z_0 is a complex number such that 1∣z0−1∣\frac{1}{|z_0 - 1|} represents the maximum value of the set {1∣z−1∣:z∈S}\left\{ \frac{1}{|z - 1|} : z \in S \right\}, then find the principal argument of 4−z0−zˉ0z0−zˉ0+2i\frac{4 - z_0 - \bar{z}_0}{z_0 - \bar{z}_0 + 2i}.

Options

A

−π2-\frac{\pi}{2}

Correct
B

π4\frac{\pi}{4}

C

π2\frac{\pi}{2}

D

3π4\frac{3\pi}{4}

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex NumbersArgand Plane

Step-by-Step Solution

To find the principal argument of the given complex expression, we break the problem down into the following logical steps:

Step 1: Understand the Geometry of Set SS

The set SS contains all complex numbers zz satisfying: ∣z−(2−i)∣≥5|z - (2 - i)| \ge \sqrt{5}

In the Argand plane, this inequality represents the region outside and on the boundary of the circle centered at C(2,−1)C(2, -1) with radius R=5R = \sqrt{5}.

Step 2: Determine the Position of z0z_0

We are given that 1∣z0−1∣\frac{1}{|z_0 - 1|} is the maximum value of {1∣z−1∣:z∈S}\left\{ \frac{1}{|z - 1|} : z \in S \right\}. This is equivalent to finding z0∈Sz_0 \in S such that the distance ∣z−1∣|z - 1| is minimized.

Let A(1,0)A(1, 0) be the point corresponding to z=1z = 1. First, let's check the distance between the center C(2,−1)C(2, -1) and point A(1,0)A(1, 0): AC=(1−2)2+(0−(−1))2=(−1)2+12=2AC = \sqrt{(1 - 2)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2}

Since AC=2<5AC = \sqrt{2} < \sqrt{5}, the point A(1,0)A(1, 0) lies inside the circle.

To minimize the distance from AA to any point zz on or outside the boundary of the circle (set SS), z0z_0 must lie on the boundary of the circle along the ray originating from CC and passing through AA.

The direction vector from CC to AA is: CA⃗=A−C=(1−2)+i(0−(−1))=−1+i\vec{CA} = A - C = (1 - 2) + i(0 - (-1)) = -1 + i

The unit vector in the direction of CA⃗\vec{CA} is: u^=−1+i∣−1+i∣=−1+i2\hat{u} = \frac{-1 + i}{|-1 + i|} = \frac{-1 + i}{\sqrt{2}}

Since z0z_0 lies on the boundary at a distance of R=5R = \sqrt{5} from C(2−i)C(2 - i) along this direction: z0=C+Ru^=(2−i)+5(−1+i2)z_0 = C + R \hat{u} = (2 - i) + \sqrt{5} \left( \frac{-1 + i}{\sqrt{2}} \right) z0=(2−52)+i(−1+52)z_0 = \left( 2 - \sqrt{\frac{5}{2}} \right) + i \left( -1 + \sqrt{\frac{5}{2}} \right)

Let z0=x0+iy0z_0 = x_0 + i y_0, where: x0=2−52x_0 = 2 - \sqrt{\frac{5}{2}} y0=−1+52y_0 = -1 + \sqrt{\frac{5}{2}}

Step 3: Simplify the Given Complex Expression

We want to evaluate the principal argument of: W=4−z0−zˉ0z0−zˉ0+2iW = \frac{4 - z_0 - \bar{z}_0}{z_0 - \bar{z}_0 + 2i}

Note that:

  • z0+zˉ0=2x0z_0 + \bar{z}_0 = 2x_0
  • z0−zˉ0=2iy0z_0 - \bar{z}_0 = 2iy_0

Substitute these into the expression: W=4−2x02iy0+2i=2(2−x0)2i(y0+1)=2−x0i(y0+1)=−i(2−x0y0+1)W = \frac{4 - 2x_0}{2iy_0 + 2i} = \frac{2(2 - x_0)}{2i(y_0 + 1)} = \frac{2 - x_0}{i(y_0 + 1)} = -i \left( \frac{2 - x_0}{y_0 + 1} \right)

Now, evaluate 2−x02 - x_0 and y0+1y_0 + 1: 2−x0=2−(2−52)=522 - x_0 = 2 - \left( 2 - \sqrt{\frac{5}{2}} \right) = \sqrt{\frac{5}{2}} y0+1=(−1+52)+1=52y_0 + 1 = \left( -1 + \sqrt{\frac{5}{2}} \right) + 1 = \sqrt{\frac{5}{2}}

Therefore: 2−x0y0+1=5/25/2=1\frac{2 - x_0}{y_0 + 1} = \frac{\sqrt{5/2}}{\sqrt{5/2}} = 1

Substituting this back gives: W=−i(1)=−iW = -i (1) = -i

Step 4: Find the Principal Argument

The principal argument of W=−iW = -i is: Arg(W)=−π2\text{Arg}(W) = -\frac{\pi}{2}

Thus, the correct option is A.

PreviousNext