If z = frac{1}{2} - 2i is such that |z+1| = a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $z = \frac{1}{2} - 2i$. Substitute $z$ into the equation $|z+1| = \alpha z + \beta(1+i)$: $|(\frac{1}{2} - 2i) + 1| = \alpha(\frac{1}{2} - 2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given $z = \frac{1}{2} - 2i$. Substitute $z$ into the equation $|z+1| = \alpha z + \beta(1+i)$: $|(\frac{1}{2} - 2i) + 1| = \alpha(\frac{1}{2} - 2i) + \beta(1+i)$ $|\frac{3}{2} - 2i| = (\frac{\alpha}{2} + \beta) + (\beta - 2\alpha)i$ Calculate the modulus on the left side: $|\frac{3}{2} - 2i| = \sqrt{(\frac{3}{2})^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Equating the real and imaginary parts: Imaginary part: $\beta - 2\alpha = 0 \implies \beta = 2\alpha$ Real part: $\frac{\alpha}{2} + \beta = \frac{5}{2}$ Substitute $\beta = 2\alpha$ into the real part equation: $\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$ $\frac{5\alpha}{2} = \frac{5}{2} \implies \alpha = 1$ Then $\beta = 2(1) = 2$. Therefore,$\alpha + \beta = 1 + 2 = 3$.

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

Similar Questions

If $\alpha, \beta$ are non-zero integers and $z=(\alpha+i \beta)(2+7 i)$ is a purely imaginary number,then the minimum value of $|z|^2$ is

Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

If $(1 + i) \cdot (1 + 2i) \dots \dots \dots (1 + ni) = x + iy$ (where $i = \sqrt{-1}$), then find the value of $(2) \cdot (5) \cdot (10) \dots \dots \dots (1 + n^2)$.

If $\log_{\tan 30^{\circ}} \left( \frac{2|z|^2 + 2|z| - 3}{|z| + 1} \right) < -2$,then:

The number of non-zero integral solutions of the equation $|1 - i|^x = 2^x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module