Help please!

How many solutions are there to Zn1=iZZ^{n-1} =i \overline{Z} and what are the solutions?

Note by Anik Mandal
3 years, 7 months ago

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Multiply both sides by z z to get zn=iz2z=1 z^n = i|z|^2 \Rightarrow |z| = 1 after considering the modulus of both sides. So, we get zn=i z^n = i which are just the nth roots of i i .

Ameya Daigavane - 3 years, 7 months ago

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How to find the nth roots of ii?

Anik Mandal - 3 years, 7 months ago

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Sorry, I should have added that. Express ii as eiπ/2e^{i\pi/2} , and let z=eiθ z = e^{i\theta} We have n roots given by,
( k k goes from 0 0 to n1 n - 1 ):
zn=eiπ/2einθ=eiπ/2nθ=2kπ+π/2θ=(4k+1)π2nzk=ei(4k+1)π2n 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 - 3 years, 7 months ago

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@Ameya Daigavane I am not too familiar with complex numbers..Can you please explain why nθ=2kπ+π2n\theta = 2k\pi+\frac{\pi}{2}

Anik Mandal - 3 years, 7 months 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 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 (ϕ \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 zz ?
Indeed, if we rotated the angle by 2π 2\pi radians then we would get the same number again. If we rotated it by 2π 2\pi again, for a total of 4π4 \pi radians, we would get z z again! So we've seen that the argument is not affected by adding a multiple of 2π2\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=rcosϕx = r \cos \phi , and y=rsinϕ y = r \sin \phi , so, z=r(cosϕ+i sinϕ)z = r( \cos \phi + i\ sin \phi )
It looks neat.
But where does eiϕe^{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,
eiϕ=(cosϕ+i sinϕ)e^{i\phi} = (\cos \phi + i\ sin \phi ) , so z=reiϕ z = r e^{i\phi}
If I'm saying z1=r1eiϕ1=z2=r2eiϕ2z_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 2π2\pi. So ϕ1ϕ2=2kπ \phi_1 - \phi_2 = 2k\pi for some integer kk .
 

This is where nθ=2kπ+π2 n\theta = 2k\pi + \frac{\pi}{2} comes from, nθ n \theta is the same angle as π2 \frac{\pi}{2} upto a multiple of 2π2\pi.

Ameya Daigavane - 3 years, 7 months 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 n1n-1?

Anik Mandal - 3 years, 7 months 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 - 3 years, 7 months ago

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@Harsh Shrivastava Why not 1 to n?

Anik Mandal - 3 years, 7 months ago

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@Anik Mandal k=0 k = 0 and k=n k = n give the same root (arguments differ by 2π2\pi).

Ameya Daigavane - 3 years, 7 months ago

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@Anik Mandal eix=cosx+isinxe^{ix} = cosx +isinx

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

Hope this helps.

Harsh Shrivastava - 3 years, 7 months ago

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@Harsh Shrivastava Yes i have understood!Thanks for answering..

Anik Mandal - 3 years, 7 months ago

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What about z=0?

Deeparaj Bhat - 3 years, 6 months ago

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

Ameya Daigavane - 3 years, 6 months ago

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@Ameya Daigavane Yeah, thanks for pointing that out.

Deeparaj Bhat - 3 years, 6 months ago

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