SATHEE: Complex Numbers 2 Question 11

Complex Numbers 2 Question 11

11.

If $| z | = 1$ and $z \neq \pm 1$ , then all the values of $\frac{z}{1 - z^{2}}$ lie on

(a) a line not passing through the origin

(b) $| z | = \sqrt{2}$

(d) the $Y$ -axis

(2007, 3M)

Show Answer

Answer:

Correct Answer: 11. (d)

Solution:

Let $z = \cos θ + i \sin θ$

$\begin{aligned} \frac{z}{1 - z^{2}} & = \frac{\cos θ + i \sin θ}{1 - (\cos 2 θ + i \sin 2 θ)} \\ = \frac{\cos θ + i \sin θ}{2 \sin^{2} θ - 2 i \sin θ \cos θ} \\ = \frac{\cos θ + i \sin θ}{- 2 i \sin θ (\cos θ + i \sin θ)} = \frac{i}{2 \sin θ} \end{aligned}$

Hence, $\frac{z}{1 - z^{2}}$ lies on the imaginary axis i.e. $Y$ -axis.

Alternate Solution

Let $E = \frac{z}{1 - z^{2}} = \frac{z}{z \bar{z} - z^{2}} = \frac{1}{\bar{z} - z}$ which is an imaginary.

Complex Numbers 2 Question 11

11.

Answer:

Alternate Solution

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Complex Numbers 2 Question 11

11.

Answer:

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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