Question

what is the argument of the complex number z = (- i - 1)/i ​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:z = \dfrac{ - 1 - i}{i}$

So, Let we first reduce this complex number to standard form.

So, on rationalizing, we get

$\rm :\longmapsto\:z = \dfrac{ - 1 - i}{i} \times \dfrac{i}{i}$

$\rm :\longmapsto\:z = \dfrac{ - i - {i}^{2} }{ {i}^{2} }$

$\rm :\longmapsto\:z = \dfrac{ - i - ( - 1) }{ - 1}$

$\rm\implies \:z = - 1 + i$

Let assume that

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

can be rewritten as

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

So, on comparing real and Imaginary parts, we get

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

and

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

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

$\rm :\longmapsto\: {(rcos\theta )}^{2} + {(rsin \theta )}^{2} = 1 + 1$

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

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

$\rm\implies \:r = \sqrt{2}$

So, on substituting the values of r in equation (1) and (2), we get

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

and

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

Since, sin is positive and cos is negative.

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

So,

$\bf\implies \:arg(z)= \dfrac{3\pi}{4}$

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Additional Information :-

Argument of complex number using Short Cut Method

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \1sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$