If z = 1/2 - 2i, is such that |z + 1| = αz + β(1 + i), i = √-1 and α, β ∈ R, then α + β is equal to

Q: Wait, why is the imaginary part on the left side zero?

The modulus of any complex number is a real number (a distance). Since $|z+1|$ is a scalar value, it has no imaginary component. Thus, $Im(LHS) = 0$.

Q: Can I just divide the equation by $z$ to find $\alpha$?

Dividing by a complex number is possible, but it complicates the equation by introducing $1/z$. It is far more elegant to expand and equate real/imaginary parts.

Q: What if $\alpha$ and $\beta$ were not real? Would the answer change?

Yes, absolutely. The constraint $\alpha, \beta \in \mathbb{R}$ is the key that allows us to equate the real and imaginary parts independently. Without it, we would have one equation with two complex variables, which is underdetermined.

Question

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

Options:

(A) -4

(B) 3

(C) 2

(D) -1

Sigma Percentile · Accepted Answer

### The Geometry of the Complex Plane

We are given a complex number $z = \frac{1}{2} - 2i$. In the complex plane, this corresponds to the point $(\frac{1}{2}, -2)$.

The problem asks us to evaluate the modulus $|z+1|$. Geometrically, adding $1$ to a complex number shifts the point one unit to the right along the real axis.

The new coordinate becomes $(\frac{1}{2} + 1, -2)$, which simplifies to $(\frac{3}{2}, -2)$. Thus, we are analyzing the complex number $\frac{3}{2} - 2i$.

### The Modulus: A Distance in Space

The modulus $|z+1|$ represents the straight-line distance from the origin to the point $\frac{3}{2} - 2i$. We calculate this using the Pythagorean theorem:

$$|z+1| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}$$

Squaring the components, we obtain:

$$|z+1| = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9+16}{4}} = \sqrt{\frac{25}{4}}$$

This simplifies to the value:

$$|z+1| = \frac{5}{2}$$

### The Diplomatic Treaty: Real vs. Imaginary

We now substitute our findings into the given equation $|z+1| = \alpha z + \beta(1+i)$. Substituting the known values, we get:

$$\frac{5}{2} = \alpha\left(\frac{1}{2} - 2i\right) + \beta(1+i)$$

Expanding the right-hand side, we distribute the constants $\alpha$ and $\beta$:

$$\frac{5}{2} = \left(\frac{\alpha}{2} - 2\alpha i\right) + (\beta + \beta i)$$

Grouping the real and imaginary parts, we obtain:

$$\frac{5}{2} = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)$$

### The Final Resolution

For two complex numbers to be equal, their real and imaginary parts must be equal independently. Since the left side has an imaginary part of $0$, we set the imaginary part of the right side to zero:

$$\beta - 2\alpha = 0 \Rightarrow \beta = 2\alpha$$

Next, we equate the real parts:

$$\frac{5}{2} = \frac{\alpha}{2} + \beta$$

Substituting $\beta = 2\alpha$ into this equation yields:

$$\frac{5}{2} = \frac{\alpha}{2} + 2\alpha = \frac{\alpha + 4\alpha}{2} = \frac{5\alpha}{2}$$

Solving for $\alpha$, we find $\alpha = 1$. Consequently, $\beta = 2(1) = 2$.

The sum of $\alpha$ and $\beta$ is:

$$\alpha + \beta = 1 + 2 = 3$$

The final answer is **3**.

Animated Solution for Mathematics - Complex Numbers: If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

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Visualized Solution

Identifying the Complex Number $z$

Calculating $z + 1$

Finding the Modulus $∣ z + 1∣$

Evaluating $∣ z + 1∣$

Setting up the Main Equation

Expanding the Right Hand Side

Grouping Real and Imaginary Parts

Equating the Imaginary Parts

Equating the Real Parts

Solving for $α$

Finding $β$ and Final Sum

Summary and Key Takeaway

The Sigma Insight: Algebraic Operations on Complex Numbers

The Geometry of the Complex Plane

The Modulus

The Diplomatic Treaty

The Final Resolution

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=21​−2i, is such that ∣z+1∣=αz+β(1+i),i=−1​ and α,β∈R, then α+β is equal to

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Visualized Solution

Identifying the Complex Number .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z+1

Finding the Modulus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z+1∣

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z+1∣