# Express the complex number $1 + i\sqrt 3 $ in modulus amplitude form.

Answer

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Hint: - You have to convert in modulus amplitude form, we know modulus of a complex number is distance of that point from origin which is equal to$ = \sqrt {{{\left( {R.p} \right)}^2} + {{\left( {I.p} \right)}^2}} $. Where ( R.P=real part and I.P=imaginary part )and amplitude of a complex number of type $a + ib$ is ${\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$.

Complete step-by-step answer:

We have the complex number $1 + i\sqrt 3 $

We can write it as by multiplying and dividing by $2$

$1 + i\sqrt 3 = 2\left( {\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)$

We know $\left( {\cos \dfrac{\pi }{3} = \dfrac{1}{2},\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}} \right)$ using these values we get,

$ = 2\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$

We know Euler’s formula ($\cos \theta + i\sin \theta = {e^{i\theta }}$) on using this formula we get,

$ = 2{e^{i\dfrac{\pi }{3}}}$ is the required modulus amplitude form.

Note: -whenever you get these types of questions the key concept of solving is you should have knowledge of how to find amplitude and modulus of a complex number. And also keep in mind Euler’s formula which is used often in these questions.

Complete step-by-step answer:

We have the complex number $1 + i\sqrt 3 $

We can write it as by multiplying and dividing by $2$

$1 + i\sqrt 3 = 2\left( {\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)$

We know $\left( {\cos \dfrac{\pi }{3} = \dfrac{1}{2},\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}} \right)$ using these values we get,

$ = 2\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$

We know Euler’s formula ($\cos \theta + i\sin \theta = {e^{i\theta }}$) on using this formula we get,

$ = 2{e^{i\dfrac{\pi }{3}}}$ is the required modulus amplitude form.

Note: -whenever you get these types of questions the key concept of solving is you should have knowledge of how to find amplitude and modulus of a complex number. And also keep in mind Euler’s formula which is used often in these questions.

Last updated date: 28th May 2023

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