Let z in mathbb{C} with Im(z) = 10 and it sat | vedclass

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

Question

(C) Let $z = x + 10i$.Given $\frac{2z - n}{2z + n} = 2i - 1$.Substituting $z = x + 10i$:$\frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1$$(2x - n) + 20

Vedclass · Accepted Answer

$n = 40$ and $Re(z) = -10$

Vedclass · Answer

(C) Let $z = x + 10i$.Given $\frac{2z - n}{2z + n} = 2i - 1$.Substituting $z = x + 10i$:$\frac{2(x + 10i) - n}{2(x + 10i) + n} = 2i - 1$$(2x - n) + 20i = (2i - 1)((2x + n) + 20i)$$(2x - n) + 20i = 2i(2x + n) + 40i^2 - (2x + n) - 20i$$(2x - n) + 20i = (2i(2x + n) - 20i) - 40 - (2x + n)$Equating real and imaginary parts:Real part: $2x - n = -40 - (2x + n)$ $\Rightarrow$ $2x - n = -40 - 2x - n$ $\Rightarrow$ $4x = -40$ $\Rightarrow$ $x = -10$.Imaginary part: $20 = 2(2x + n) - 20$ $\Rightarrow$ $40 = 2(2x + n)$ $\Rightarrow$ $20 = 2x + n$.Substituting $x = -10$ into $20 = 2x + n$:$20 = 2(-10) + n$ $\Rightarrow$ $20 = -20 + n$ $\Rightarrow$ $n = 40$.Thus, $n = 40$ and $Re(z) = -10$.

Let $z \in \mathbb{C}$ with $Im(z) = 10$ and it satisfies $\frac{2z - n}{2z + n} = 2i - 1$ for some natural number $n$. Then

Similar Questions

Let $x$ and $y$ be real numbers such that $50 \left(\frac{2x}{1 + 3i} - \frac{y}{1 - 2i}\right) = 31 + 17i$, where $i = \sqrt{-1}$. Then the value of $10(x - 3y)$ is:

The real value of $\alpha$ for which $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is

The imaginary part of $\frac{(1 + i)^2}{2 - i}$ is

$\left(\frac{1+i}{1-i}\right)^{228} = $

The statement $(a + ib) < (c + id)$ is true for

Vedclass Test Series

Exam Paper Generator

Online Exam Module