Question

If z=cos theta+isin theta then (1+z)/(1-z) is equal to

Answer 1

Answer:

$(1 + z)(1 - z) = 2sin \theta(sin \theta - 2icos \theta)$

Step-by-step explanation:

Given that-

$z = cos \theta+isin \theta$

(1 + z)(1 - z) =?

We have-

(1 + z)(1 - z)

$= (1+cos \theta+isin \theta)[1-(cos \theta+isin \theta)]$

We know that-

$(a-b)(a+b) = a^{2} - b^{2}$

$= 1^{2} - (cos \theta+isin \theta)^{2}$

$= 1 -(cos^{2} \theta + i^{2}sin^{2} \theta + 2icos \theta sin \theta)$

$= 1 -cos^{2} \theta -i^{2} sin^{2} \theta - 2icos \theta sin \theta$

But, $i^{2} = -1$

$= 1 - cos^{2} \theta + sin^{2} \theta -2isin \theta cos \theta$

But, $1 - cos^{2} \theta = sin^{2} \theta$

$= sin^{2} \theta + sin^{2} \theta - 2isin \theta cos \theta$

$= 2sin^{2} \theta - 2icos \theta sin \theta$

$= 2sin \theta(sin \theta - 2icos \theta)$

$(1 + z)(1 - z) = 2sin \theta(sin \theta - 2icos \theta)$

Answer 2

Answer:

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