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Hint-These types of questions can be solved by using the formula of modulus and argument of

the complex number.

Given complex number is

$z = - 1 - i\sqrt 3 $

Now we know that the general form of complex number is

$z = x + iy$

Now comparing the above two we get,

$x = - 1$ and ${\text{ }}y = - \sqrt 3 $

Now let’s find the modulus of the complex number.

We know that the modulus of a complex number is $\left| z \right|$

$\left| z \right| = \sqrt {{x^2} + {y^2}} $

Now putting the value of $x$ and $y$ we get,

$

\left| z \right| = \sqrt {{{( - 1)}^2} + {{( - \sqrt 3 )}^2}} \\

\left| z \right| = \sqrt {1 + 3} \\

\left| z \right| = \sqrt 4 \\

\left| z \right| = 2 \\

$

Therefore, the modulus of a given complex number is $2$.

Now let’s find the argument of the complex number.

Now we know that the general form of complex number is

$z = x + iy$

Let $x$ be $r\cos \theta $ and$y$ be $r\sin \theta $ where $r$ is the modulus of the complex number.

Now putting the values of $x$ and $y$ in $z$ we get,

$z = r\cos \theta + ir\sin \theta $

Now comparing the above two we get,

$ - 1 - i\sqrt 3 {\text{ }} = r\cos \theta + ir\sin \theta $

Now, comparing the real parts we get,

${\text{ - 1 = }}r\cos \theta $

Now, putting the value of $r$ in the above equation we get,

$

{\text{ - 1 = 2}}\cos \theta \\

{\text{or }}\cos \theta = \dfrac{{ - 1}}{2}{\text{ }} \\

$

Similarly, compare the imaginary parts and put the value of $r$ we get,

$

- \sqrt 3 = 2\sin \theta \\

{\text{or }}\sin \theta = - \dfrac{{\sqrt 3 }}{2} \\

$

Hence, $\sin \theta = - \dfrac{{\sqrt 3 }}{2}{\text{ and }}\cos \theta = \dfrac{{ - 1}}{2}{\text{ }}$

or$\theta {\text{ = }}{60^ \circ }$

Now we can clearly see that the values of both $\sin \theta $ and$\cos \theta $ are negative.

And we know that they both are negative in ${3^{rd}}$ quadrant.

Therefore, the argument is in ${3^{rd}}$ quadrant.

Argument${\text{ = - (18}}{0^ \circ } - \theta {\text{)}}$

$

{\text{ = - (18}}{0^ \circ } - \theta {\text{)}} \\

= {\text{ - (18}}{0^ \circ } - {60^ \circ }{\text{)}} \\

{\text{ = - (12}}{{\text{0}}^ \circ }{\text{)}} \\

{\text{ = - 12}}{0^ \circ } \\

$

Now converting it in $\pi $ form we get,

$

= - {120^ \circ } \times \dfrac{\pi }{{{{180}^ \circ }}} \\

= - \dfrac{{2\pi }}{3} \\

$

Hence, the argument of complex number is $ - \dfrac{{2\pi }}{3}$

Note- Whenever we face such types of questions the key concept is that we simply compare the given

complex number with its general form and then find the value of $x$ and $y$ and put it in the formula

of modulus and argument of the complex number

the complex number.

Given complex number is

$z = - 1 - i\sqrt 3 $

Now we know that the general form of complex number is

$z = x + iy$

Now comparing the above two we get,

$x = - 1$ and ${\text{ }}y = - \sqrt 3 $

Now let’s find the modulus of the complex number.

We know that the modulus of a complex number is $\left| z \right|$

$\left| z \right| = \sqrt {{x^2} + {y^2}} $

Now putting the value of $x$ and $y$ we get,

$

\left| z \right| = \sqrt {{{( - 1)}^2} + {{( - \sqrt 3 )}^2}} \\

\left| z \right| = \sqrt {1 + 3} \\

\left| z \right| = \sqrt 4 \\

\left| z \right| = 2 \\

$

Therefore, the modulus of a given complex number is $2$.

Now let’s find the argument of the complex number.

Now we know that the general form of complex number is

$z = x + iy$

Let $x$ be $r\cos \theta $ and$y$ be $r\sin \theta $ where $r$ is the modulus of the complex number.

Now putting the values of $x$ and $y$ in $z$ we get,

$z = r\cos \theta + ir\sin \theta $

Now comparing the above two we get,

$ - 1 - i\sqrt 3 {\text{ }} = r\cos \theta + ir\sin \theta $

Now, comparing the real parts we get,

${\text{ - 1 = }}r\cos \theta $

Now, putting the value of $r$ in the above equation we get,

$

{\text{ - 1 = 2}}\cos \theta \\

{\text{or }}\cos \theta = \dfrac{{ - 1}}{2}{\text{ }} \\

$

Similarly, compare the imaginary parts and put the value of $r$ we get,

$

- \sqrt 3 = 2\sin \theta \\

{\text{or }}\sin \theta = - \dfrac{{\sqrt 3 }}{2} \\

$

Hence, $\sin \theta = - \dfrac{{\sqrt 3 }}{2}{\text{ and }}\cos \theta = \dfrac{{ - 1}}{2}{\text{ }}$

or$\theta {\text{ = }}{60^ \circ }$

Now we can clearly see that the values of both $\sin \theta $ and$\cos \theta $ are negative.

And we know that they both are negative in ${3^{rd}}$ quadrant.

Therefore, the argument is in ${3^{rd}}$ quadrant.

Argument${\text{ = - (18}}{0^ \circ } - \theta {\text{)}}$

$

{\text{ = - (18}}{0^ \circ } - \theta {\text{)}} \\

= {\text{ - (18}}{0^ \circ } - {60^ \circ }{\text{)}} \\

{\text{ = - (12}}{{\text{0}}^ \circ }{\text{)}} \\

{\text{ = - 12}}{0^ \circ } \\

$

Now converting it in $\pi $ form we get,

$

= - {120^ \circ } \times \dfrac{\pi }{{{{180}^ \circ }}} \\

= - \dfrac{{2\pi }}{3} \\

$

Hence, the argument of complex number is $ - \dfrac{{2\pi }}{3}$

Note- Whenever we face such types of questions the key concept is that we simply compare the given

complex number with its general form and then find the value of $x$ and $y$ and put it in the formula

of modulus and argument of the complex number

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