Question

Find the Modules & argument of the complex Number 4√3 + 4i.​

Answer 1

Given : Complex number is $4\sqrt{3}+4i$

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To find : Modulus and argument of the given complex number.

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Solution :

Let's say the given complex number be,

• $z = 4\sqrt{3}+4i$

And Assume that,

• $y = {\sf{Im}}(z)=4$
• $x = {\sf{Re}}(z) =4\sqrt{3}$

Modulus of a complex number is given by,

• $\overline{z} = r = \sqrt{x^{2} + {y}^{2} }$

Substitute the values of $x$ and $y$

$\implies r = \sqrt{x^{2} + {y}^{2} }$

$\implies r = \sqrt{(4 \sqrt{3}) ^{2} + {(4)}^{2} }$

$\implies r = \sqrt{48 + 16 }$

$\implies r = \sqrt{64 }$

$\implies r = 8$

So the modulus of the given complex number is $8$.

Argument of a complex number in first quadrant is defined as,

• $\theta = \tan^-\left|\dfrac{y}{x}\right|$

Here,

• $\theta$ = Argument of the complex number.

• $y$ = Imaginary part of complex number.

• $x$ = Real part of the complex number

By substituting the values of $x$ and $y$ , we get :

$\implies \theta = \tan^-\Big|\dfrac{y}{x}\Big|$

$\implies \theta = \tan^-\bigg|\dfrac{4}{4 \sqrt{3} }\bigg|$

$\implies \theta = \tan^-\bigg|\dfrac{ \not4}{ \not4 \sqrt{3} }\bigg|$

$\implies \theta = \tan^-\bigg|\dfrac{ 1}{ \sqrt{3} }\bigg|$

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

Hence the argument of the given complex number is $\dfrac{\pi}{6}$

Additional information :-

• Argument of complex number in 1st quadrant is given by $\theta$ where $\theta$ is measured anticlockwise.

• Argument of complex number in 2nd quadrant is given by $\pi - \theta$ where $\theta$ is measured anticlockwise.

• Argument of complex number in 3rd quadrant is given by $\theta - \pi$ and $\theta$ is measured clockwise .

• Argument of complex number in 4th quadrant is given by $- \theta$ where $\theta$ is measured clockwise.

Answer 2

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\:4 \sqrt{3} + 4i$

Let we assume that

$\rm :\longmapsto\:4 \sqrt{3} + 4i = r(cos\theta + isin\theta )$

where,

$\red{ \rm :\longmapsto\:r \: is \: modulus \: or \: length \: of \: complex \: number}$

and

$\red{ \rm :\longmapsto\:\theta \: is \: argument \: of \: complex \: number}$

So,

$\rm :\longmapsto\:4 \sqrt{3} + 4i = rcos\theta + i \: r \: sin\theta$

On comparing, Real part on both sides, we get

$\red{ \rm :\implies\:\boxed{ \sf{ \:rcos\theta = 4 \sqrt{3}}}} - - - (1)$

and

On comparing Imaginary parts on both sides, we get

$\red{ \rm :\implies\:\boxed{ \sf{ \:rsin\theta = 4 }}} - - - (2)$

On squaring equation (1) and (2) and adding we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta + {r}^{2} {sin}^{2}\theta = 48 + 16$

$\rm :\longmapsto\: {r}^{2}( {cos}^{2}\theta+ {sin}^{2}\theta)= 64$

$\rm :\longmapsto\: {r}^{2}(1)= 64$

$\rm :\longmapsto\: {r}^{2}= 64$

$\bf\implies \:r = 8 \: \: as \: r \: \cancel < \: 0 - - - - (3)$

Now, on substituting r = 8 in equation (2) and (3), we get

$\rm :\longmapsto\:8cos\theta = 4 \sqrt{3}$

$\bf\implies \:cos\theta = \dfrac{ \sqrt{3} }{2} - - - - (4)$

and

$\rm :\longmapsto\:8sin\theta = 4$

$\bf\implies \:sin\theta = \dfrac{1}{2} - - - - (5)$

As,

$\rm :\longmapsto\:sin\theta > 0 \: \: and \: \: cos\theta > 0$

$\rm :\implies\:\theta \: lies \: in \: {1}^{st} \: quadrant$

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

Hence,

$\: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \: |z| = 8}}} \: \: \: \green{and} \: \: \: \red{ \boxed{ \sf{ \:arg(z) = \dfrac{\pi}{6} \: }}}$

Additional Information :-

$\red{ \boxed{ \sf{ \: {i}^{2} \: = \: - 1 \: \: }}}$

$\red{ \boxed{ \sf{ \: {i}^{3} \: = \: -i \: \: }}}$

$\red{ \boxed{ \sf{ \: {i}^{4} \: = \: 1 \: \: }}}$

$\red{ \boxed{ \sf{ \: { \omega}^{3} = 1}}}$

$\red{ \boxed{ \sf{ \:1 + \omega \: + \: { \omega}^{2} = 0}}}$