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

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

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TopNewestMultiply 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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– Deeparaj Bhat · 6 months ago

What about z=0?Log in to reply

– Ameya Daigavane · 6 months ago

Smooth observation, missed that during the multiplication. Then we also require \( n > 1 \), though.Log in to reply

– Deeparaj Bhat · 6 months ago

Yeah, thanks for pointing that out.Log in to reply

– Anik Mandal · 6 months, 2 weeks ago

How to find the nth roots of \(i\)?Log in to reply

( \( k \) 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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– Anik Mandal · 6 months, 2 weeks ago

I am not too familiar with complex numbers..Can you please explain why \(n\theta = 2k\pi+\frac{\pi}{2}\)Log in to reply

When you look at the complex plane, you see this:

the complex plane

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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– Anik Mandal · 6 months, 2 weeks ago

That's was really helpful Sir!Thanks a lot! Sir one last doubt : Why was it k from 0 to \(n-1\)?Log in to reply

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

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– Anik Mandal · 6 months, 2 weeks ago

Why not 1 to n?Log in to reply

– Ameya Daigavane · 6 months, 2 weeks ago

\( k = 0 \) and \( k = n \) give the same root (arguments differ by \(2\pi\)).Log in to reply

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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– Anik Mandal · 6 months, 2 weeks ago

Yes i have understood!Thanks for answering..Log in to reply