Question

If $\mathcal{Z}=\frac{\sqrt{3}}{2}+\frac{i}{2}$, Find $Arg\left(\mathcal{-Z}\right)$.​

Answer 1

We have a complex Number :-

$\mathcal{Z}=\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}$

We know that every complex Number can be written in its polar form .The polar form of a complex Number is given as follows in general:-

$\mathcal{Z}=r(cos\theta +isin\theta)$

Now we have :-

$r(cos\theta +isin\theta)=\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}$

On comaprison we get the following :-

$rcos\theta=\dfrac{\sqrt{3}}{2}\\ \\ \\r^2cos^2\theta=\dfrac{3}{4}$

Also we have :-

$rsin\theta=\dfrac{1}{2}\\ \\ \\r^2sin^2\theta=\dfrac{1}{4}$

Now :-

$r^2(cos^2\theta+sin^2\theta)=\dfrac{3}{4}+\dfrac{1}{4}\\ \\ \\r^2=\dfrac{4}{4}=1\\ \\ \\r=1$

now:-

$\dfrac{rsin\theta}{rcos\theta}=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ \\ \\tan\theta=\dfrac{1}{\sqrt{3}}$

we know that :-

$tan\theta=\dfrac{y}{x}$

Hence ,the value of tan theta is positive .

This means that theta lies in First Quadrant.

now:-

$tan\theta=tan(\dfrac{\pi}{6})$

Hence:-

$arg(\mathcal{Z})=\dfrac{\pi}{6}$

Now,

We know that :-

$arg(-z)-arg(z)=\pm \pi\\ \\ \\ \\arg(\mathcal{-Z})=\pm \pi+arg(\mathcal{Z})\\ \\ \\ \\arg(\mathcal{-Z})=\pm \pi+\dfrac{\pi}{6}$

Answer 2

Answer:

hope it is helpful to you.