Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of z3 + 3z2

Solution & Explanation

Core Logic

The given equations represent two circles in the complex plane: Circle 1: Center

C₁(6, 0)

, radius

r₁ = 5

Circle 2: Center

C₂(-2, 6)

, radius

r₂ = 5

Distance between the centers $C₁$ $C_1$ and $C₂$ $C_2$:

C₁C₂ = \sqrt((-2 - 6)² + (6 - 0)²) = \sqrt(64 + 36) = 10

Notice that $C₁C₂ = r₁ + r₂ = 5 + 5 = 10$ $C_1C_2 = r_1 + r_2 = 5 + 5 = 10$. This means the two circles touch each other externally at exactly one point.

Step 1: Finding z

Since the circles touch externally, the common point $z$ $z$ is the midpoint of the line segment joining the centers $C₁$ $C_1$ and $C₂$ $C_2$.

z = (6 + (-2))/(2) + i (0 + 6)/(2)

$z = 2 + 3i$ $z = 2 + 3i$

Step 2: Simplifying the Polynomial

We have $z = 2 + 3i$ $z = 2 + 3i$. $z - 2 = 3i$ $z - 2 = 3i$ Squaring both sides: $(z - 2)² = -9$ $(z - 2)^2 = -9$

z² - 4z + 4 = -9 z² - 4z + 13 = 0

$z² = 4z - 13$ $z^2 = 4z - 13$

Step 3: Evaluating the Expression

We need to evaluate $z³ + 3z² - 15z + 141$ $z^3 + 3z^2 - 15z + 141$. First, find $z³$ $z^3$:

z³ = z \cdot z² = z(4z - 13) = 4z² - 13z

Substitute $z² = 4z - 13$ $z^2 = 4z - 13$ again:

z³ = 4(4z - 13) - 13z = 16z - 52 - 13z = 3z - 52

Now substitute $z³$ $z^3$ and $z²$ $z^2$ into the target expression:

(3z - 52) + 3(4z - 13) - 15z + 141

= 3z - 52 + 12z - 39 - 15z + 141

= (3z + 12z - 15z) + (-52 - 39 + 141)

$= 0 + 50 = 50$ $= 0 + 50 = 50$

Pattern Recognition

When given two complex distance modulus equations $|z-z₁|=r₁$ $|z-z_1|=r_1$ and $|z-z₂|=r₂$ $|z-z_2|=r_2$, always check the distance between centers $|z₁ - z₂|$ $|z_1 - z_2|$. If it exactly equals $r₁ + r₂$ $r_1 + r_2$, the single unique solution is the section formula midpoint.

Chapter Mix

Class 11 Mathematics: Complex Numbers and Quadratic Equations

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument

α

denotes the number of solutions of

|1 - i|^x = 2^x

and

β = ((|z|)/( (z)))

, where

z = (π)/(4) (1 + i)⁴ ( 1 - \sqrt(π) i\sqrt(π) + i + \sqrt(π) - i1 + \sqrt(π) i), i = \sqrt(-1)

, then the distance of the point

(α, β)

from the line

4x - 3y = 7

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

|1 - i|^x = 2^x (\sqrt(2))^x = 2^x 2x/2 = 2^x

This implies $(x)/(2) = x x = 0$ $\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so $α = 1$ $\alpha = 1$.

Step 1: Simplify complex number z

(1+i)⁴ = ((1+i)²)² = (1 + i² + 2i)² = (2i)² = -4

Thus, $z = -π ( (1-\sqrt(π)i)(\sqrt(π)-i)π + 1 + (\sqrt(π)-i)(1-\sqrt(π)i)1 + π )$

Step 2: Simplify Bracket

Let's expand the terms directly: $z = (π)/(4)(-4) [ \sqrt(π) - π i - i - \sqrt(π)π + 1 + \sqrt(π) - i - π i - \sqrt(π)1 + π ]$

= -π [ (-i(π+1))/(π+1) + (-i(π+1))/(π+1) ]

= -π [ -i - i ] = 2π i

Step 3: Find beta

For $z = 2π i$ $z = 2\pi i$: $|z| = 2π$ $|z| = 2\pi$ and $(z) = (π)/(2)$ $\arg(z) = \frac{\pi}{2}$.

β = (|z|)/( (z)) = (2π)/(π/2) = 4

Step 4: Distance from Line

Distance of point $(α, β) = (1, 4)$ $(\alpha, \beta) = (1, 4)$ from the line $4x - 3y - 7 = 0$ $4x - 3y - 7 = 0$:

D = |4(1) - 3(4) - 7|\sqrt(4² + (-3)²) = (|4 - 12 - 7|)/(5) = (|-15|)/(5) = 3

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

Let $z$ $z$ be a complex number such that $|z - 6| = 5$ $|z - 6| = 5$ and $|z + 2 - 6i| = 5$ $|z + 2 - 6i| = 5$. Then the value of $z³ + 3z² - 15z + 141$ $z^{3} + 3z^{2} - 15z + 141$ is equal to

Solution & Explanation

Core Logic

Step 1: Finding z

Step 2: Simplifying the Polynomial

Step 3: Evaluating the Expression

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers and Quadratic Equations Questions — jee_main_2026_28_january_morning

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Let z$z$ be a complex number such that |z - 6| = 5$|z - 6| = 5$ and |z + 2 - 6i| = 5$|z + 2 - 6i| = 5$. Then the value of z³ + 3z² - 15z + 141$z^{3} + 3z^{2} - 15z + 141$ is equal to

Solution & Explanation

Core Logic

Step 1: Finding z

Step 2: Simplifying the Polynomial

Step 3: Evaluating the Expression

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers and Quadratic Equations Questions — jee_main_2026_28_january_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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Let $z$ $z$ be a complex number such that $|z - 6| = 5$ $|z - 6| = 5$ and $|z + 2 - 6i| = 5$ $|z + 2 - 6i| = 5$. Then the value of $z³ + 3z² - 15z + 141$ $z^{3} + 3z^{2} - 15z + 141$ is equal to