SATHEE: Complex Numbers 5 Question 1

Complex Numbers 5 Question 1

1. If $z$ and $w$ are two complex numbers such that $| z w | = 1$ and $\arg (z) - \arg (w) = \frac{π}{2}$ , then

(2019 Main, 10 April II)

(a) $\bar{z} w = - i$

(b) $z \bar{w} = \frac{1 - i}{\sqrt{2}}$

(d) $z \bar{w} = \frac{- 1 + i}{\sqrt{2}}$

Show Answer

Answer:

Correct Answer: 1. (a)

Solution:

It is given that, there are two complex numbers $z$ and $w$ , such that $| z w | = 1$ and $\arg (z) - \arg (w) = π / 2$ $∴ | z | | w | = 1$ $[∵ | z_{1} z_{2} | = | z_{1} | | z_{2} |]$ and $\arg (z) = \frac{π}{2} + \arg (w)$

Let $| z | = r$ , then $| w | = \frac{1}{r}$

and let $\arg (w) = θ$ , then $\arg (z) = \frac{π}{2} + θ$

So, we can assume

$z = r e^{i (π / 2 + θ)}$

$[∵$ if $z = x + i y$ is a complex number, then it can be written as $z = r e^{i θ}$ where, $r = | z | and θ = \arg (z)]$

$and w = \frac{1}{r} e^{i θ}$

Now, $\bar{z} \cdot w = r e^{- i (π / 2 + θ)} \cdot \frac{1}{r} e^{i θ}$

$= e^{i (- π / 2 - θ + θ)} = e^{- i (π / 2)} = - i [∵ e^{- i θ} = \cos θ - i \sin θ]$

and

$\begin{aligned} z \bar{w} = r e^{i (π / 2 + θ)} \cdot \frac{1}{r} e^{- i θ} \\ = e^{i (π / 2 + θ - θ)} = e^{i (π / 2)} = i \end{aligned}$

Complex Numbers 5 Question 1

1. If $z$ and $w$ are two complex numbers such that $| z w | = 1$ and $\arg (z) - \arg (w) = \frac{π}{2}$ , then

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Complex Numbers 5 Question 1

1. If z and w are two complex numbers such that |zw|=1 and arg⁡(z)−arg⁡(w)=π2, then

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

1. If $z$ and $w$ are two complex numbers such that $| z w | = 1$ and $\arg (z) - \arg (w) = \frac{π}{2}$ , then