JEE Advanced 2019

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: Let $S$ be the set of all complex numbers $z$ satisfying $∣ z - 2 + i ∣ \geq 5$ . If the complex number $z_{0}$ is such that $\frac{1}{∣ z _{0} - 1∣}$ is the maximum of the set ${\frac{1}{∣ z - 1∣} : z \in S}$ , then the principal argument of $\frac{4 - z _{0} - z ˉ _{0}}{z _{0} - z ˉ _{0} + 2 i}$ is

Q: I got $W = -i$. Is that correct? How do I find the principal argument of $-i$?

Absolutely correct! If $W = -i$, you are looking at a point on the negative imaginary axis. The principal argument is the angle measured from the positive real axis. Since $-i$ is at an angle of $270^\circ$ (or $-90^\circ$) counter-clockwise, the principal argument is $-\frac{\pi}{2}$. You've nailed it!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the complex plane. The inequality

∣ z - (2 - i) ∣ \geq 5

defines a region. Can you identify the center and radius of the circle that forms the boundary of this region? Where does the point

P (1, 0)

lie relative to this circle?

Hint 2

Hint 2: You are asked to maximize

\frac{1}{∣ z _{0} - 1∣}

. This is equivalent to minimizing the distance

∣ z_{0} - 1∣

between the point

z_{0}

and the fixed point

P (1, 0)

. Since

P

is inside the circle, where must the point

z_{0}

on the boundary be located to minimize this distance?

Hint 3

Hint 3: The point

z_{0}

must lie on the line segment connecting the center of the circle

C (2, - 1)

and the point

P (1, 0)

. Find the equation of this line. Once you have the relation between

x_{0}

and

y_{0}

from this line, substitute it into the expression

\frac{4 - z _{0} - z ˉ _{0}}{z _{0} - z ˉ _{0} + 2 i}

to simplify it.

Visualized Solution

The Region $S$

$∣ z - (2 - i) ∣ \geq 5$
Center: $C (2, - 1)$
Radius: $R = 5$
Set $S$ is the exterior and boundary of this circle.

Maximizing the Expression

Maximize $\frac{1}{∣ z - 1∣}$ for $z \in S$
Equivalent to minimizing $∣ z - 1∣$
$∣ z - 1∣$ is the distance from $z$ to $P (1, 0)$

Position of Point $P$

Distance $C P = ∣1 - (2 - i) ∣ = ∣ - 1 + i ∣$
$C P = (- 1)^{2} + 1^{2} = 2$
Since $2 < 5$ , $P$ is strictly inside the circle.

Locating $z_{0}$

Minimum distance occurs on the boundary $∣ z - 2 + i ∣ = 5$ .
The points $C$ , $P$ , and $z_{0}$ must be collinear.
$z_{0}$ is the intersection of ray $C P$ and the circle.

Equation of Line $C P$

Slope of $C P$ : $m = \frac{0 - ( - 1 )}{1 - 2} = - 1$
Equation: $y - 0 = - 1 (x - 1)$
$⟹ x + y = 1$

Coordinates of $z_{0}$

Let $z_{0} = x_{0} + i y_{0}$
Since $z_{0}$ lies on the line $x + y = 1$ :
$x_{0} + y_{0} = 1$

The Target Expression

Target: $W = \frac{4 - ( z _{0} + z ˉ _{0} )}{( z _{0} - z ˉ _{0} ) + 2 i}$
Recall: $z_{0} + \overset{z}{ˉ}_{0} = 2 x_{0}$
Recall: $z_{0} - \overset{z}{ˉ}_{0} = 2 i y_{0}$

Substituting Real and Imaginary Parts

$W = \frac{4 - 2 x _{0}}{2 i y _{0} + 2 i}$
$W = \frac{2 ( 2 - x _{0} )}{2 i ( y _{0} + 1 )}$

Simplifying the Expression

$W = \frac{2 - x _{0}}{i ( y _{0} + 1 )}$
Since $\frac{1}{i} = - i$ :
$W = - i (\frac{2 - x _{0}}{y _{0} + 1})$

Analyzing the Signs

From the graph, $z_{0}$ is to the left of $x = 1$ and above $y = 0$ .
$x_{0} < 1 ⟹ 2 - x_{0} > 0$
$y_{0} > 0 ⟹ y_{0} + 1 > 0$
Therefore, $\frac{2 - x _{0}}{y _{0} + 1} > 0$

Principal Argument

$W = - k i$ where $k > 0$
The principal argument of $- k i$ is $- \frac{π}{2}$
Conclusion: The required argument is $- \frac{π}{2}$ .

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

The set

S

is defined by the inequality

∣ z - 2 + i ∣ \geq 5

. This represents the exterior and the boundary of a circle centered at

C (2, - 1)

with a radius

R = 5

Our objective is to maximize the expression

\frac{1}{∣ z _{0} - 1∣}

. To maximize this fraction, we must minimize the denominator

∣ z_{0} - 1∣

, which represents the distance between a point

z_{0} \in S

and the fixed point

P (1, 0)

The Hunt for the Minimum

First, we determine the position of

P

relative to the circle. The distance

C P

is calculated as:

C P = ∣ (1 + 0 i) - (2 - i) ∣ = ∣ - 1 + i ∣ = (- 1)^{2} + (1)^{2} = 2

Since

2 < 5

, the point

P

lies strictly inside the circle. The point

z_{0}

on the boundary closest to

P

must lie on the line passing through

C

and

P

The Algebraic Unmasking

The slope

m

of the line passing through

C (2, - 1)

and

P (1, 0)

is:

m = \frac{0 - ( - 1 )}{1 - 2} = \frac{1}{- 1} = - 1

Using the point-slope form at

P (1, 0)

, the equation of the line is

y - 0 = - 1 (x - 1)

, which simplifies to

x + y = 1

. Thus, for any point

z_{0} = x_{0} + i y_{0}

on this line, we have the constraint

x_{0} + y_{0} = 1

We now evaluate the expression

W = \frac{4 - ( z _{0} + z ˉ _{0} )}{( z _{0} - z ˉ _{0} ) + 2 i}

. Using the identities

z_{0} + \overset{z}{ˉ}_{0} = 2 x_{0}

and

z_{0} - \overset{z}{ˉ}_{0} = 2 i y_{0}

, we substitute these into the expression:

W = \frac{4 - 2 x _{0}}{2 i y _{0} + 2 i} = \frac{2 ( 2 - x _{0} )}{2 i ( y _{0} + 1 )} = \frac{2 - x _{0}}{i ( y _{0} + 1 )}

The Final Revelation

Since

\frac{1}{i} = - i

, the expression becomes

W = - i (\frac{2 - x _{0}}{y _{0} + 1})

. Given the constraint

x_{0} + y_{0} = 1

, we can substitute

y_{0} = 1 - x_{0}

into the denominator:

y_{0} + 1 = (1 - x_{0}) + 1 = 2 - x_{0}

Substituting this back into our expression for

W

W = - i (\frac{2 - x _{0}}{2 - x _{0}}) = - i

The value of the expression is $- i$ . The principal argument of

- i

is $- \frac{π}{2}$ .

Common Traps & FAQs

Wait, why do we reject the negative root or the other side of the circle? How do we know

z_{0}

is the closest point?

Great question! In geometry, the shortest distance from a point inside a circle to its boundary is always along the normal—which, for a circle, is the radial line passing through the center. By extending the line segment

C P

until it hits the boundary, we are guaranteed to find the point

z_{0}

that minimizes the distance

∣ z_{0} - 1∣

. Any other point on the circle would be further away.

Why can't we just use energy conservation or some other physics principle here?

While physics is beautiful, this is a pure complex analysis problem! We are dealing with the geometry of the Argand plane. The 'energy' here is the distance function. We are essentially finding the 'equilibrium' point where the distance is minimized. Stick to the geometric properties of complex numbers—they are your best tools here.

I got

W = - i

. Is that correct? How do I find the principal argument of

- i

Absolutely correct! If

W = - i

, you are looking at a point on the negative imaginary axis. The principal argument is the angle measured from the positive real axis. Since

- i

is at an angle of

27 0^{\circ}

(or

- 9 0^{\circ}

) counter-clockwise, the principal argument is

- \frac{π}{2}

. You've nailed it!

Select Answer:

Visualized Solution

The Region $S$

Maximizing the Expression

Position of Point $P$

Locating $z_{0}$

Equation of Line $C P$

Coordinates of $z_{0}$

The Target Expression

Substituting Real and Imaginary Parts

Simplifying the Expression

Analyzing the Signs

Principal Argument

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Hunt for the Minimum

The Algebraic Unmasking

The Final Revelation

Similar Questions

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 2∣ \leq 1}$ and $S_{2} = {z \in C : z (1 + i) + \overset{z}{ˉ} (1 - i) \geq 4}$ . Then, the maximum value of $∣ z - \frac{5}{2} ∣^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to :

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let $S = {z \in C : ∣ z - 1∣ = 1 and (2 - 1) (z + \overset{z}{ˉ}) - i (z - \overset{z}{ˉ}) = 22}$ . Let $z_{1}, z_{2} \in S$ be such that $∣ z_{1} ∣ = max_{z \in S} ∣ z ∣$ and $∣ z_{2} ∣ = min_{z \in S} ∣ z ∣$ . Then $∣ 2 z_{1} - z_{2} ∣^{2}$ equals :

Let z be those complex number which satisfy $∣ z + 5∣ \leq 4$ and $z (1 + i) + \overset{z}{ˉ} (1 - i) \geq - 10$ , $i = - 1$ If the maximum value of $∣ z + 1 ∣^{2}$ is $α + β 2$ , then the value of $(α + β)$ is

Let $z$ be a complex number such that $∣ z ∣ = 1$ . If $\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle $∣ z - (1 + 2 i) ∣ = 1$ is:

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .

Select Answer:

Visualized Solution

The Region .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Maximizing the Expression

Position of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Locating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z0​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CP

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z0​

The Target Expression

Substituting Real and Imaginary Parts

Simplifying the Expression

Analyzing the Signs

Principal Argument

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Hunt for the Minimum

The Algebraic Unmasking

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be the complex number satisfying ∣z−5∣≤3 and having maximum positive principal argument. Then 34​5iz+165z−12​​2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number such that ∣z∣=1. If k+zˉ2+k2z​=kz,k∈R, then the maximum distance of k+ik2 from the circle ∣z−(1+2i)∣=1 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for the complex numbers z satisfying ∣z−2−2i∣≤1, the maximum value of ∣3iz+6∣ is attained at a+ib, then a+b is equal to .

The Region $S$

Position of Point $P$

Locating $z_{0}$

Equation of Line $C P$

Coordinates of $z_{0}$

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let $z$ be a complex number such that $∣ z ∣ = 1$ . If $\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle $∣ z - (1 + 2 i) ∣ = 1$ is:

If for the complex numbers $z$ satisfying $∣ z - 2 - 2 i ∣ \leq 1$ , the maximum value of $∣3 i z + 6∣$ is attained at $a + ib$ , then $a + b$ is equal to .