Question

find modulus and argument of. -1-i​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:z = - 1 - i$

Let assume that,

$\bf :\longmapsto\: - 1 - i = r(cos\theta + isin\theta ) - - - - (1)$

where,

r is Modulus of complex number

θ is argument of complex number.

Now,

$\bf :\longmapsto\: - 1 - i = rcos\theta + i \: rsin\theta$

On comparing, we get

$\rm :\longmapsto\:rcos\theta = - 1 - - - (2)$

and

$\rm :\longmapsto\:rsin\theta = - 1 - - - (3)$

On squaring equation (2) and (3) we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta = 1 - - - (4)$

and

$\rm :\longmapsto\: {r}^{2} {sin}^{2}\theta = 1 - - - (5)$

On adding equation (4) and (5), we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta + {r}^{2} {sin}^{2}\theta = 1 + 1$

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

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

$\bf\implies \:r = \sqrt{2} \: \: \: as \: r > 0$

On substituting the value of r in equation (2) and (3), we get

$\rm :\longmapsto\:cos\theta = - \: \dfrac{1}{ \sqrt{2} }$

and

$\rm :\longmapsto\:sin\theta = - \: \dfrac{1}{ \sqrt{2} }$

As cosθ < 0 and sinθ < 0

$\bf\implies \:\theta \: \in \: {3}^{rd} \: quadrant$

$\rm :\implies\:\theta = - \pi + \dfrac{\pi}{4}$

$\rm :\implies\:\theta = \dfrac{ - 4\pi + \pi}{4}$

$\rm :\implies\:\theta = \dfrac{ - 3\pi}{4}$

$\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:Hence-\begin{cases} &\bf{Modulus = \sqrt{2} } \\ \\ &\bf{Argument = - \: \dfrac{3\pi}{4} } \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

Argument of a complex number

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf Re(z)&\sf \: Im(z)&\sf \:arg(z)\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf > 0&\sf > 0&\sf\theta \\\\\sf < 0&\sf > 0&\sf\pi - \theta \\\\\sf < 0 &\sf < 0&\sf\ - \pi + \theta \\\\\sf > 0&\sf < 0&\sf - \theta \\\\ \frac{\qquad}{}&\frac{\qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$