Name: Basic Geometrical Concepts of Complex Number Plane
Uploaded: 2019-07-19
Duration: 5 min 49 s
Description: Before studying the geometrical applications of complex numbers, it is essential to recall the basic geometrical concepts of the complex number plane. Let's...

Question

Throughout this question the use of a calculator is not permitted.

The complex numbers $w$ and $z$ satisfy the relation $w = \frac{z + i}{i z + 2} .$

Given instead that $w = z$ and the real part of $z$ is negative, find $z,$ giving your answer in the form $x + iy,$ where $x$ and $y$ are real.

Accepted Answer

Given the complex numbers $w$ and $z$ satisfy the relation $w = \frac{z + i}{i z + 2}$ that $w = z$ and the real part of $z$ is negative.

$\Rightarrow z = \frac{z + i}{i z + 2} (w = z)$

Let's take $z = - a + b i$

$\Rightarrow - a + b i = \frac{- a + b i + i}{i (- a + b i) + 2}$

$\Rightarrow (- a + b i) \times (i (- a + b i) + 2) = - a + (1 + b) i$

$w$

$\Rightarrow (- a + b i) \times (- a i - b + 2) = - a + (1 + b) i$ where $i^{2} = - 1$

$\Rightarrow a^{2} i + a b - 2 a - a b i^{2} - b^{2} i + 2 b i = - a + (1 + b) i$

$\Rightarrow (a^{2} - b^{2}) i + a b - 2 a + a b + 2 b i = - a + (1 + b) i$

$\Rightarrow (a^{2} - b^{2} + 2 b) i + 2 a b - 2 a = - a + (1 + b) i$

Equating real and imaginary parts

$\Rightarrow 2 a b - 2 a = - a$ and $a^{2} - b^{2} + 2 b = 1 + b$

$\Rightarrow 2 a b - a = 0$

$\Rightarrow a \times (2 b - 1) = 0$

$\Rightarrow a = 0$ and $b = \frac{1}{2}$

Here $a = 0$ is possible because it is given real part is negative so substituting $b = \frac{1}{2}$ in $a^{2} - b^{2} + 2 b = 1 + b$

$\Rightarrow a^{2} - {\frac{1}{2}}^{2} + 2 \times \frac{1}{2} = 1 + \frac{1}{2}$

$\Rightarrow a^{2} - \frac{1}{4} = \frac{1}{2}$

$\Rightarrow a^{2} = \frac{1}{2} + \frac{1}{4} = \frac{2 + 1}{4} = \frac{3}{4}$

$\Rightarrow a = \pm \frac{\sqrt{3}}{2}$

Given real part is negative so $a = - \frac{\sqrt{3}}{2}$

Therefore, $z = - \frac{\sqrt{3}}{2} + \frac{1}{2} i .$

Throughout this question the use of a calculator is not permitted.

The complex numbers $w$ and $z$ satisfy the relation $w = \frac{z + i}{i z + 2} .$

Given instead that $w = z$ and the real part of $z$ is negative, find $z,$ giving your answer in the form $x + iy,$ where $x$ and $y$ are real.

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Throughout this question the use of a calculator is not permitted. The complex numbers w and z satisfy the relation w=z+iiz+2. Given instead that w=z and the real part of z is negative, find z, giving your answer in the form x+iy, where x and y are real.

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Throughout this question the use of a calculator is not permitted.

The complex numbers $w$ and $z$ satisfy the relation $w = \frac{z + i}{i z + 2} .$

Given instead that $w = z$ and the real part of $z$ is negative, find $z,$ giving your answer in the form $x + iy,$ where $x$ and $y$ are real.