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How many solutions are there to \[Z^{n-1} =i \overline{Z}\] and what are the solutions?

Note by Anik Mandal
6 months, 2 weeks 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 \( i \). Ameya Daigavane · 6 months, 2 weeks ago

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@Ameya Daigavane What about z=0? Deeparaj Bhat · 6 months ago

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@Deeparaj Bhat Smooth observation, missed that during the multiplication. Then we also require \( n > 1 \), though. Ameya Daigavane · 6 months ago

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@Ameya Daigavane Yeah, thanks for pointing that out. Deeparaj Bhat · 6 months ago

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@Ameya Daigavane How to find the nth roots of \(i\)? Anik Mandal · 6 months, 2 weeks ago

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@Anik Mandal 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 \( n - 1 \)):
\( z^n = e^{i\pi/2} \\ \Rightarrow e^{in\theta} = e^{i\pi/2} \\ \Rightarrow n\theta = 2k\pi + \pi/2 \\ \Rightarrow \theta = \frac{(4k+ 1)\pi}{2n} \\ \Rightarrow z_k = e^{\frac{i(4k+ 1)\pi}{2n}} \) Ameya Daigavane · 6 months, 2 weeks ago

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@Ameya Daigavane I am not too familiar with complex numbers..Can you please explain why \(n\theta = 2k\pi+\frac{\pi}{2}\) Anik Mandal · 6 months, 2 weeks ago

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@Anik Mandal Sure!
When you look at the complex plane, you see this:

the complex plane

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 \(2\pi\). Ameya Daigavane · 6 months, 2 weeks ago

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@Ameya Daigavane That's was really helpful Sir!Thanks a lot! Sir one last doubt : Why was it k from 0 to \(n-1\)? Anik Mandal · 6 months, 2 weeks ago

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@Anik Mandal A polynomial of degree n has only n zeroes.

Thus k is from 0 to (n-1). Harsh Shrivastava · 6 months, 2 weeks ago

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@Harsh Shrivastava Why not 1 to n? Anik Mandal · 6 months, 2 weeks ago

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@Anik Mandal \( k = 0 \) and \( k = n \) give the same root (arguments differ by \(2\pi\)). Ameya Daigavane · 6 months, 2 weeks ago

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@Anik Mandal \(e^{ix} = cosx +isinx\)

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

Hope this helps. Harsh Shrivastava · 6 months, 2 weeks ago

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@Harsh Shrivastava Yes i have understood!Thanks for answering.. Anik Mandal · 6 months, 2 weeks ago

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