Question

-root3-i
Express the following
complex Nurnbess in modulus
amplitude fom​

Answer 1

Given :-

• Complex number is -√3 - i

To find :-

• Modulus amplitude form of complex number.

Solution :-

Modulus amplitude of complex number z is given by,

$:\implies z = |z| \bigg( \cos \theta + i \sin \theta\bigg)$

Here,

• | z | = Modulus of complex number.
• θ = Angle made by complex number on argànd plane from origin.

Let's assume that,

• $x = -\sqrt 3 (\sf\:Real\:part \:of\:z)$
• $y = -1 \:\sf(Imaginary\; part\:of\:z)$

Modulus of a complex number is given by,

$: \implies |z| = \sqrt{ Re{(z)}^{2} + Im {(z)}^{2} }$

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

$: \implies |z| = \sqrt{ {(- \sqrt{3} } )^{2} + {( - 1)}^{2} }$

$: \implies |z| = \sqrt{ 3 + 1}$

$: \implies |z| = \sqrt{ 4}$

$: \implies |z| =\pm 2$

Now we have to find the angle θ

$: \implies \tan \alpha = \left | \dfrac{y}{x} \right|$

$: \implies \tan \alpha = \left | \dfrac{ - 1}{ - \sqrt{3} } \right|$

$: \implies \tan \alpha = \left | \dfrac{ 1}{ \sqrt{3} } \right|$

$: \implies \tan \alpha = \dfrac{ 1}{ \sqrt{3} }$

Since tan A is equals to 1/√3 on π/6,

$: \implies \alpha = \dfrac{\pi}{6}$

Since both $x$ and $y$ are negative, the given complex number will lie in the 3rd quadrant on argànd plane.

Angle of 3rd quadrant is given by,

$: \implies \theta = - (\pi - \alpha )$

$: \implies \theta = \alpha - \pi$

$: \implies \theta = \dfrac{\pi}{6} - \pi$

$: \implies \theta = - \dfrac{ 5\pi}{6}$

This θ is called the argument of complex number.

Now put value of θ and α in general modulus amplitude form.

$: \implies |z| \left( \cos \theta + i \sin \theta \right)$

$: \implies \pm 2 \left( \cos \dfrac{-5\pi}{6}+ i \sin \dfrac{-5\pi}{6} \right)$

Refer to the attachments for more details.