How many solutions are there to $Z^{n-1} =i \overline{Z}$ and what are the solutions? Note by Anik Mandal
4 years, 8 months ago

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Multiply both sides by $z$ to get $z^n = i|z|^2 \Rightarrow |z| = 1$ after considering the modulus of both sides. So, we get $z^n = i$ which are just the nth roots of $i$.

- 4 years, 8 months ago

How to find the nth roots of $i$?

- 4 years, 8 months ago

Sorry, I should have added that. Express $i$ as $e^{i\pi/2}$, and let $z = e^{i\theta}$ We have n roots given by,
( $k$ goes from $0$ to $n - 1$):
$z^n = e^{i\pi/2} \\ \Rightarrow e^{in\theta} = e^{i\pi/2} \\ \Rightarrow n\theta = 2k\pi + \pi/2 \\ \Rightarrow \theta = \frac{(4k+ 1)\pi}{2n} \\ \Rightarrow z_k = e^{\frac{i(4k+ 1)\pi}{2n}}$

- 4 years, 8 months ago

I am not too familiar with complex numbers..Can you please explain why $n\theta = 2k\pi+\frac{\pi}{2}$

- 4 years, 8 months ago

Sure!
When you look at the complex plane, you see this: the complex plane Each point in this plane determines a unique complex number.
For a point, the x coordinate determines the real part, and the y coordinate determines the imaginary part. In the figure, we have $z = x + iy$.
This is called the Cartesian representation of a complex number. We can also represent complex numbers in another form, called the polar form.
In the polar form, each number is represented by two parameters, its modulus (r in the figure, the distance from the origin) and its argument ($\phi$, the angle shown in the figure). By varying these parameters, we can reach every complex number.
But if you think about it, how can the argument take a unique value for a specific $z$?
Indeed, if we rotated the angle by $2\pi$ radians then we would get the same number again. If we rotated it by $2\pi$ again, for a total of $4 \pi$ radians, we would get $z$ again! So we've seen that the argument is not affected by adding a multiple of $2\pi$.
Keep this in mind whenever we think about arguments of complex numbers.

We need to find a way to relate the two representations. Trigonometry is the way!
Looking at the right-angled triangle in the figure,
$x = r \cos \phi$, and $y = r \sin \phi$, so, $z = r( \cos \phi + i\ sin \phi )$
It looks neat.
But where does $e^{i\phi}$ come from? The answer is here, Euler's formula.
This is the reason why complex numbers are so powerful. With this, we have,
$e^{i\phi} = (\cos \phi + i\ sin \phi )$, so $z = r e^{i\phi}$
If I'm saying $z_1 = r_1e^{i\phi_1} = z_2 = r_2e^{i\phi_2}$, I mean their moduli are equal, but their arguments could differ by a multiple of $2\pi$. So $\phi_1 - \phi_2 = 2k\pi$ for some integer $k$.

This is where $n\theta = 2k\pi + \frac{\pi}{2}$ comes from, $n \theta$ is the same angle as $\frac{\pi}{2}$ upto a multiple of $2\pi$.

- 4 years, 8 months ago

That's was really helpful Sir!Thanks a lot! Sir one last doubt : Why was it k from 0 to $n-1$?

- 4 years, 8 months ago

A polynomial of degree n has only n zeroes.

Thus k is from 0 to (n-1).

- 4 years, 8 months ago

Why not 1 to n?

- 4 years, 8 months ago

$k = 0$ and $k = n$ give the same root (arguments differ by $2\pi$).

- 4 years, 8 months ago

$e^{ix} = cosx +isinx$

Put $x=pi/2,5pi/2,9pi/2 \cdots ,2kpi +pi/2$, all will yeild i.

Hope this helps.

- 4 years, 8 months ago

Yes i have understood!Thanks for answering..

- 4 years, 8 months ago

- 4 years, 8 months ago

Smooth observation, missed that during the multiplication. Then we also require $n > 1$, though.

- 4 years, 8 months ago

Yeah, thanks for pointing that out.

- 4 years, 8 months ago