If z=(1/2)-2i, is such that |z+1|=alpha z+beta(1+i), i=sqrt(-1) and alpha,beta R, then alpha+beta is equal to

Solution & Explanation

Related Formula

|x + iy| = \sqrt(x² + y²)

Two complex numbers are equal if and only if their real and imaginary parts are respectively equal.

Core Logic

Given $z = (1)/(2) - 2i$ $z = \frac{1}{2} - 2i$. Calculate $|z + 1|$ $|z + 1|$:

|z + 1| = | ((1)/(2) - 2i) + 1 |

= | (3)/(2) - 2i |

= \sqrt(((3)/(2))² + (-2)²) = \sqrt((9)/(4) + 4) = \sqrt((25)/(4)) = (5)/(2)

Now substitute $z$ $z$ and $|z + 1|$ $|z + 1|$ into the original equation:

(5)/(2) = α((1)/(2) - 2i) + β(1 + i)

Expand and group real and imaginary components on the RHS:

(5)/(2) = ((α)/(2) - 2α i) + (β + β i)

(5)/(2) = ((α)/(2) + β) + i(β - 2α)

Step 1: Equate Parts

By equating the real and imaginary parts from both sides, we get a system of linear equations:

Imaginary part:

0 = β - 2α \Rightarrow β = 2α

Real part:

(5)/(2) = (α)/(2) + β

Substitute $β = 2α$ $\beta = 2\alpha$ into the real part equation:

(5)/(2) = (α)/(2) + 2α

(5)/(2) = (5α)/(2)

$α = 1$ $\alpha = 1$

Using $α = 1$ $\alpha = 1$, find $β$ $\beta$:

β = 2(1) = 2

Step 2: Final Calculation

Calculate the final requested value:

α + β = 1 + 2 = 3

Pattern Recognition

Equating complex parts reduces single complex equations into two simultaneous linear equations. Treat $|z+1|$ $|z+1|$ strictly as a scalar magnitude and parse directly into algebraic components.

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

If $z=(1)/(2)-2i$ $z=\frac{1}{2}-2i$, is such that $|z+1|=α z+β(1+i)$ $|z+1|=\alpha z+\beta(1+i)$, $i=\sqrt(-1)$ $i=\sqrt{-1}$ and $α,βin R$ $\alpha,\beta\in R$, then $α+β$ $\alpha+\beta$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Equate Parts

Step 2: Final Calculation

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_2024_29_jan_morning

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If z=(1)/(2)-2i$z=\frac{1}{2}-2i$, is such that |z+1|=α z+β(1+i)$|z+1|=\alpha z+\beta(1+i)$, i=√(-1)$i=\sqrt{-1}$ and α,βin R$\alpha,\beta\in R$, then α+β$\alpha+\beta$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Equate Parts

Step 2: Final Calculation

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_2024_29_jan_morning

Select Your Target Exam

Full Mock Test Hub

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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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If $z=(1)/(2)-2i$ $z=\frac{1}{2}-2i$, is such that $|z+1|=α z+β(1+i)$ $|z+1|=\alpha z+\beta(1+i)$, $i=\sqrt(-1)$ $i=\sqrt{-1}$ and $α,βin R$ $\alpha,\beta\in R$, then $α+β$ $\alpha+\beta$ is equal to