Question

Solve question given in image​

Answer 1

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Answer 2

Answer:

Option (C) is correct

Step-by-step explanation:

We are given the complex number,

${ \longrightarrow z = \dfrac{(1+i\sqrt3)(\cos\theta+i\sin\theta)}{2(1-i)(\cos\theta-i\sin\theta)}}$

We have to find the modulus of this complex number.

We know that modulus of complex number say x + iy is given by,

$\bullet \: \: \: |z| = \sqrt{ {x}^{2} + {y}^{2} }$

Also, the modulus can be distributed over the complex terms.

So,

${ \longrightarrow |z| = \dfrac{ | 1+i\sqrt3 || \cos\theta+i\sin\theta | }{2 | 1-i || \cos\theta-i\sin\theta | }}$

${ \longrightarrow |z| = \dfrac{ \sqrt{ {1}^{2} + { \sqrt{3} }^{2} } \times \sqrt{( \cos \theta)^{2} + ( \sin \theta)^{2} } }{2 \times \sqrt{ {1}^{2} + { (- 1)}^{2}} \times \sqrt{( \cos \theta)^{2} + ( \sin \theta)^{2} }}}$

${ \longrightarrow |z| = \dfrac{ \sqrt{1 + 3 } \times \sqrt{ \cos^{2} \theta + \sin ^{2} \theta } }{2 \times \sqrt{ 1 + 1} \times \sqrt{ \cos ^{2} \theta + \sin^{2} \theta }}}$

${ \longrightarrow |z| = \dfrac{ \sqrt{4 } \times \sqrt{ \cos^{2} \theta + \sin ^{2} \theta } }{2 \times \sqrt{2} \times \sqrt{ \cos ^{2} \theta + \sin^{2} \theta }}}$

${ \longrightarrow |z| = \dfrac{ \sqrt{4 } }{2 \times \sqrt{2} }}$

${ \longrightarrow |z| = \dfrac{ 2}{2 \times \sqrt{2} }}$

${ \longrightarrow |z| = \dfrac{1}{ \sqrt{2} }}$

Hence option (c) is correct.