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 \( i \).

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 \( n - 1 \)):
\( z^n = e^{i\pi/2} \\ \Rightarrow e^{in\theta} = e^{i\pi/2} \\ \Rightarrow n\theta = 2k\pi + \pi/2 \\ \Rightarrow \theta = \frac{(4k+ 1)\pi}{2n} \\ \Rightarrow z_k = e^{\frac{i(4k+ 1)\pi}{2n}} \)

@Anik Mandal
–
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 \(2\pi\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \( i \).

Log in to reply

What about z=0?

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

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 \( n - 1 \)):

\( z^n = e^{i\pi/2} \\ \Rightarrow e^{in\theta} = e^{i\pi/2} \\ \Rightarrow n\theta = 2k\pi + \pi/2 \\ \Rightarrow \theta = \frac{(4k+ 1)\pi}{2n} \\ \Rightarrow z_k = e^{\frac{i(4k+ 1)\pi}{2n}} \)

Log in to reply

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 \(2\pi\).

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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

Hope this helps.

Log in to reply

Log in to reply