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Sorry, I should have added that. Express i as eiπ/2, and let z=eiθ
We have n roots given by,
( k goes from 0 to n−1): zn=eiπ/2⇒einθ=eiπ/2⇒nθ=2kπ+π/2⇒θ=2n(4k+1)π⇒zk=e2ni(4k+1)π
@Anik Mandal
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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 (ϕ, 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 2π radians then we would get the same number again. If we rotated it by 2π again, for a total of 4π radians, we would get z again! So we've seen that the argument is not affected by adding a multiple of 2π.
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=rcosϕ, and y=rsinϕ, so, z=r(cosϕ+isinϕ)
It looks neat.
But where does eiϕ 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+isinϕ), so z=reiϕ
If I'm saying z1=r1eiϕ1=z2=r2eiϕ2, I mean their moduli are equal, but their arguments could differ by a multiple of 2π. So ϕ1−ϕ2=2kπ for some integer k.
This is where nθ=2kπ+2π comes from, nθ is the same angle as 2π upto a multiple of 2π.
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Multiply both sides by z to get zn=i∣z∣2⇒∣z∣=1 after considering the modulus of both sides. So, we get zn=i which are just the nth roots of i.
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How to find the nth roots of i?
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Sorry, I should have added that. Express i as eiπ/2, and let z=eiθ We have n roots given by,
( k goes from 0 to n−1):
zn=eiπ/2⇒einθ=eiπ/2⇒nθ=2kπ+π/2⇒θ=2n(4k+1)π⇒zk=e2ni(4k+1)π
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nθ=2kπ+2π
I am not too familiar with complex numbers..Can you please explain whyLog in to reply
When you look at the complex plane, you see this:
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 (ϕ, 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 2π radians then we would get the same number again. If we rotated it by 2π again, for a total of 4π radians, we would get z again! So we've seen that the argument is not affected by adding a multiple of 2π.
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=rcosϕ, and y=rsinϕ, so, z=r(cosϕ+i sinϕ)
It looks neat.
But where does eiϕ 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), so z=reiϕ
If I'm saying z1=r1eiϕ1=z2=r2eiϕ2, I mean their moduli are equal, but their arguments could differ by a multiple of 2π. So ϕ1−ϕ2=2kπ for some integer k.
This is where nθ=2kπ+2π comes from, nθ is the same angle as 2π upto a multiple of 2π.
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n−1?
That's was really helpful Sir!Thanks a lot! Sir one last doubt : Why was it k from 0 toLog in to reply
Thus k is from 0 to (n-1).
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k=0 and k=n give the same root (arguments differ by 2π).
eix=cosx+isinx
Put x=pi/2,5pi/2,9pi/2⋯,2kpi+pi/2, all will yeild i.
Hope this helps.
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What about z=0?
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Smooth observation, missed that during the multiplication. Then we also require n>1, though.
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