MHB Finding roots of a complex number

[Drain Brain]

I'm trying to solve this problem and got stuck.

Find the roots of $\sqrt{-j}$

converting $0-j$ into polar form

$r=\sqrt{0^2-1^2}=1$

$\theta=\tan^{-1}\left(\frac{-1}{0}\right)$ I got stuck on this part. please help.

 

[Prove It]

I'm trying to solve this problem and got stuck.

Find the roots of $\sqrt{-j}$

converting $0-j$ into polar form

$r=\sqrt{0^2-1^2}=1$

$\theta=\tan^{-1}\left(\frac{-1}{0}\right)$ I got stuck on this part. please help.

Have you not even thought to plot -j on an Argand Diagram? This would make finding the argument (angle) unbelievably easy...

 

[Drain Brain]

Have you not even thought to plot -j on an Argand Diagram? This would make finding the argument (angle) unbelievably easy...

I have not considered that.

I can see when I graph $0-j$ it is lying on y-axis donward making an angle of 270 deg.

 

[Prove It]

I have not considered that.

I can see when I graph $0-j$ it is lying on y-axis donward making an angle of 270 deg.

Except that the angles are always measured in radians and the principal argument is measured in the domain $\displaystyle \begin{align*} \theta \in \left( -\pi, \pi \right) \end{align*}$.

You will also need to get a GENERAL angle (since the arguments are repeated every $\displaystyle \begin{align*} 2\pi \end{align*}$ units).

 

[Drain Brain]

Except that the angles are always measured in radians and the principal argument is measured in the domain $\displaystyle \begin{align*} \theta \in \left( -\pi, \pi \right) \end{align*}$.

You will also need to get a GENERAL angle (since the arguments are repeated every $\displaystyle \begin{align*} 2\pi \end{align*}$ units).

the principle angle should be $-\frac{\pi}{2}$ correct?

what do you mean by "general angle"?

 

[Prove It]

the principle angle should be $-\frac{\pi}{2}$ correct?

what do you mean by "general angle"?

Meaning that in general, the angle is of the form $\displaystyle \begin{align*} -\frac{\pi}{2} + 2\,\pi\,n \end{align*}$, where n is an integer.

 

[Deveno]

What you are really doing is solving the complex quadratic:

$z^2 + j = 0$.

If we write:

$z = r(\cos\theta + j\sin\theta)$, we have:

$z^2 = r^2(\cos2\theta + j\sin2\theta) = 0 + j(-1)$.

It follows that $r = 1$, and that:

$\cos2\theta = 0$
$\sin2\theta = -1$.

Now there are an infinite number of $\theta$'s that would work, but we only need to find 2 that correspond to two different complex numbers.

 

[Drain Brain]

What you are really doing is solving the complex quadratic:

$z^2 + j = 0$.

If we write:

$z = r(\cos\theta + j\sin\theta)$, we have:

$z^2 = r^2(\cos2\theta + j\sin2\theta) = 0 + j(-1)$.

It follows that $r = 1$, and that:

$\cos2\theta = 0$
$\sin2\theta = -1$.

Now there are an infinite number of $\theta$'s that would work, but we only need to find 2 that correspond to two different complex numbers.

from your solution how do I find $\theta$?

I'm thinking of taking the inverse of
$\cos2\theta = 0$
$\sin2\theta = -1$.

is that correct?

 

[Deveno]

from your solution how do I find $\theta$?

I'm thinking of taking the inverse of
$\cos2\theta = 0$
$\sin2\theta = -1$.

is that correct?

Well, the inverse trig functions on a calculator, let's say, would give you:

$\cos^{-1}(0) = \dfrac{\pi}{2}$
$\sin^{-1}(-1) = -\dfrac{\pi}{2}$

which doesn't give you a consistent answer. BE VERY CAREFUL WITH INVERSE TRIG FUNCTIONS. The sine and cosine functions are NOT one-to-one, they are PERIODIC.

But...where does (0,-1) lie on the unit circle? There are many "angles" one can associate with that point (add $2\pi$ to anyone of them, and you get another one), but there is only the ONE point. So...pick an angle that works, and divide it by two. That's one of your square roots...what's the other one?