# Find the roots of the complex number ##(-1+i)^\frac {1}{3}##

• chwala
In summary: I have in my post ###5##.@Mark44 that is exactly what I used in post ##5##.You have an error - or at least a typo - in the exponent of the following. From Post #5: ##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3 ##That should be: ##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac
chwala
Gold Member
Homework Statement
Kindly see attached...I just want to understand why for the case; ##(-1+i)^\frac {1}{3}## they divided by ##3## and not multiply ?
Relevant Equations
Complex numbers
Kindly see attached...I just want to understand why for the case; ##(-1+i)^\frac {1}{3}## they divided by ##3## when working out the angles...
Am assuming they used;
##(\cos x + i \sin x)^n = \cos nx + i \sin nx## and here, we require ##n## to be positive integers...unless I am not getting it...sorry using phone to type ...I will put this into context later...

For ##z^5 = -32## No issue here...the steps ,that is division by ##5## is clear.

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Write cis ##x## as abbreviation for ##(\cos x + i\sin x)##.
Assume we can write ##(-1+i)^{\frac13}## as ##z=r\,\textrm{cis}\,\theta##.
From de Moivre we have
$$-1+i=z^3=(r\,\textrm{cis}\,\theta)^3 = r^3\textrm{cis}\,3\theta$$
So the argument (angle) for ##-1+i## in the complex plane is three times the angle of ##z##. So to get the angle of ##z## we need to divide the angle of ##-1+i## by 3.

topsquark and chwala
andrewkirk said:
Write cis ##x## as abbreviation for ##(\cos x + i\sin x)##.
Assume we can write ##(-1+i)^{\frac13}## as ##z=r\,\textrm{cis}\,\theta##.
From de Moivre we have
$$-1+i=z^3=(r\,\textrm{cis}\,\theta)^3 = r^3\textrm{cis}\,3\theta$$
So the argument (angle) for ##-1+i## in the complex plane is three times the angle of ##z##. So to get the angle of ##z## we need to divide the angle of ##-1+i## by 3.
@andrewkirk thanks let me go through your post later... at times these things can mix someone ..
.cheers mate

Polar form of complex number,
$$-1+i= \sqrt{2}\ e^{3/4\ \pi\ i}$$, is easy to handle power 1/3. Dividing phase by 3 and multiplying cubic roots of 1,
$$(-1+i)^{1/3}= \sqrt[6]{2}\ e^{1/4\ \pi i + 2/3\ n\pi i}$$
where n=-1,0,1.

chwala and topsquark
chwala said:
@andrewkirk thanks let me go through your post later... at times these things can mix someone ..
.cheers mate
Aaargh can see now...was blind I like your approach @andrewkirk ...I will post my understanding later...now clear . Thanks @anuttarasammyak

...This way is clear to me;

Let ##z^3=(-1+i)##

... then, ##(-1+i)^\frac{1}{3}=z##

In polar form,

##z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3##

## =2^\frac{1}{6}\left[\cos(\dfrac{\dfrac {3π}{4}+2kπ}{3})+ i \sin (\dfrac{\dfrac {3π}{4}+2kπ}{3})\right]##​
Supposing we have ##z=(-1+i)^3## how would we deal with this? binomial theorem? ...let me check this out...​

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anuttarasammyak said:
Polar form of complex number,
$$-1+i= \sqrt{2}\ e^{3/4\ \pi\ i}$$, is easy to handle power 1/3. Dividing phase by 3 and multiplying cubic roots of 1,
$$(-1+i)^{1/3}= \sqrt[6]{2}\ e^{1/4\ \pi i + 2/3\ n\pi i}$$
where n=-1,0,1.
I think error here...supposed to be;

$$-1+i= \sqrt{2}\ e^{3/4\ \pi\ i+2\ n\pi i}$$,

You are right but not an error because
$$e^{2n\pi i}=1$$

chwala said:
Aaargh can see now...was blind I like your approach @andrewkirk ...I will post my understanding later...now clear . Thanks @anuttarasammyak

...This way is clear to me;

Let ##z^3=(-1+i)##

... then, ##(-1+i)^\frac{1}{3}=z##

In polar form,

##z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3##
The above is wrong. You have mixed up the representations of z and ##z^3##.
Much simpler:
Let ##z = -1 + i##
Converting to polar form, we have ##z = 2^{1/2} cis(\frac{3\pi}4 + 2k\pi)## where cis(A) means cos(A) + i sin(A).
Then ##z^{1/3} = 2^{1/6} cis(\frac{\pi}4 + k\frac{2\pi}3), k = 0, 1, 2##

Notice that there are 3 cube roots of z. All have modulus (magnitude) ##2^{1/6}##. Their arguments (angles) are ##\pi/4, \pi/4 + 2\pi/3##, and ##\pi/4 + 4\pi/3## respectively.
chwala said:
## =2^\frac{1}{6}\left[\cos(\dfrac{\dfrac {3π}{4}+2kπ}{3})+ i \sin (\dfrac{\dfrac {3π}{4}+2kπ}{3})\right]##​

Supposing we have ##z=(-1+i)^3## how would we deal with this? binomial theorem? ...let me check this out...​
Start with ##z = -1 + i = \sqrt 2 cis(3\pi/4)##. Now raise z to the power 3.

Mark44 said:
The above is wrong. You have mixed up the representations of z and ##z^3##.
Much simpler:
Let ##z = -1 + i##
Converting to polar form, we have ##z = 2^{1/2} cis(\frac{3\pi}4 + 2k\pi)## where cis(A) means cos(A) + i sin(A).
Then ##z^{1/3} = 2^{1/6} cis(\frac{\pi}4 + k\frac{2\pi}3), k = 0, 1, 2##

Notice that there are 3 cube roots of z. All have modulus (magnitude) ##2^{1/6}##. Their arguments (angles) are ##\pi/4, \pi/4 + 2\pi/3##, and ##\pi/4 + 4\pi/3## respectively.

Start with ##z = -1 + i = \sqrt 2 cis(3\pi/4)##. Now raise z to the power 3.
Looks like something wrong then in post ##2## in defining ##z##.

chwala said:
Looks like something wrong then in post ##2## in defining ##z##.
No, I don't see anything wrong there.
andrewkirk said:
Assume we can write ##(-1+i)^{\frac13}## as ##z=r\,\textrm{cis}\,\theta##.
From de Moivre we have
$$-1+i=z^3=(r\,\textrm{cis}\,\theta)^3 = r^3\textrm{cis}\,3\theta$$
The difference between what andrewkirk wrote and what I wrote is that we defined z differently. Andrew is defining z as ##(-1 + i)^{1/3}##. I am defining z as -1 + i. It doesn't matter what you start with as long as your work is consistent with how you defined things.

The error in your work in post #5 is here:
Let ##z^3=(-1+i)##
... then, ##(-1+i)^\frac{1}{3}=z##

In polar form,
##z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3##
That exponent of 3 at the far right should not be there.

If ##z = r(\cos(\theta) + i\sin(\theta)) = r cis(\theta)##,
then one value of ##z^{1/n}## is ##r^{1/n} cis(\theta/n)##.

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Mark44 said:
No, I don't see anything wrong there.
The difference between what andrewkirk wrote and what I wrote is that we defined z differently. Andrew is defining z as ##(-1 + i)^{1/3}##. I am defining z as -1 + i. It doesn't matter what you start with as long as your work is consistent with how you defined things.
@Mark44 that is exactly what I used in post ##5##.

chwala said:
@Mark44 that is exactly what I used in post ##5##.
No it's not. Take another look at my post #10. I added to it.

chwala said:
@Mark44 that is exactly what I used in post ##5##.
You have an error - or at least a typo - in the exponent of the following.

From Post #5:
##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3 ##

That should be:
##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^{1/3} ##

chwala
chwala said:
..This way is clear to me; Let ##z^3=(-1+i)##
It seems a lot simpler to do it this way:
Let ##z = -1 + i = \sqrt 2(\cos(3\pi/4 + i\sin(3\pi/4) = \sqrt 2 cis(3\pi/4)##
We can add multiples of ##2\pi## to the argument (angle), so the last expression is equal to ##\sqrt 2 cis(3\pi/4 + 2k\pi)##, with ##k \in \mathbb Z##.

Then ##z^n = (2^{1/2})^n cis\left[ n(3\pi/4 + 2k\pi)\right]##
The above is not limited to integer values of n. n can take on rational values, such as 1/3.

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SammyS said:
You have an error - or at least a typo - in the exponent of the following.

From Post #5:
##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^3 ##

That should be:
##\displaystyle z=2^\frac{1}{6}\left[\cos (\dfrac{3π}{4}+2kπ)+ i \sin (\dfrac{3π}{4}+2kπ)\right]^{1/3} ##
My confusion is here;

is it correct therefore to say

##[\cos θ + i \sin θ]^n = cos n θ + i \sin nθ##

and

##[\cos θ + i \sin θ]^\frac{1}{n} = \cos n θ + i \sin nθ##

Mark44 said:
It seems a lot simpler to do it this way:
Let ##z = -1 + i = \sqrt 2(\cos(3\pi/4 + i\sin(3\pi/4) = \sqrt 2 cis(3\pi/4)##
We can add multiples of ##2\pi## to the argument (angle), so the last expression is equal to ##\sqrt 2 cis(3\pi/4 + 2k\pi)##, with ##k \in \mathbb Z##.

Then ##z^n = (2^{1/2})^n cis\left[ n(3\pi/4 + 2k\pi)\right]##
The above is not limited to integer values of n. n can take on rational values, such as 1/3.
Thanks Mark, do we treat/substitute ##n=3## and ##n=\dfrac{1}{3}## the same way? that is specifically on,

##z^n = ...cis\left[ \frac{3\pi/4 + 2k\pi}n\right]##

Last edited by a moderator:
chwala said:
My confusion is here;

is it correct therefore to say

##[\cos θ + i \sin θ]^n = \cos n θ + i \sin nθ##

and

##[\cos θ + i \sin θ]^\frac{1}{n} = \cos n θ + i \sin nθ##
Yes.

No.

In that order.

chwala said:
Thanks Mark, do we treat/substitute ##n=3## and ##n=\dfrac{1}{3}## the same way? that is specifically on,
##z^n = ...cis\left[ \frac{3\pi/4 + 2k\pi}n\right]##
No. I claim credit for some of your confusion. The above, which is part of something I wrote but have since corrected, should not have the angle divided by 3. The angle should have been multiplied by 3. For the problem in this thread, we have ##z = -1 + i##, so ##|z| = \sqrt 2##, and ##\theta = \frac{3\pi}4##.

Then, ##z^3 = |z|^3cis( 3(\theta))##
and ##z^{1/3} = |z|^{1/3}cis(\frac 1 3(\theta + 2k\pi))##, with |z| and ##\theta## as above.
For positive integral powers, you don't need to include that extra ##k2\pi## term - raising a number to a positive integer power gets you just a single value.
For roots, you do need to include the multiples of ##2\pi## in order to get all of the roots. If you don't include that extra part, you'll get only the principal n-th root.

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Mark44 said:
No. I claim credit for some of your confusion. The above, which is part of something I wrote but have since corrected, should not have the angle divided by 3. The angle should have been multiplied by 3. For the problem in this thread, we have ##z = -1 + i##, so ##|z| = \sqrt 2##, and ##\theta = \frac{3\pi}4##.

Then, ##z^3 = |z|^3cis( 3(\theta))##
and ##z^{1/3} = |z|^{1/3}cis(\frac 1 3(\theta + 2k\pi))##, with |z| and ##\theta## as above.
For positive integral powers, you don't need to include that extra ##k2\pi## term - raising a number to a positive integer power gets you just a single value.
For roots, you do need to include the multiples of ##2\pi## in order to get all of the roots. If you don't include that extra part, you'll get only the principal n-th root.
That was my initial argument in post ##1## when I asked why are we not multiplying...anyway I'll need to go through the entire thread again...cheers man!

Lalalala...I think I now got it! I will post my understanding later... cheers guys!

I can summarise as follows with your indulgence; if we are solving;

1. ##z^n=x^n## where ##x## is say an integer value i.e from ##z=x+iy## with ##y=0## and ##n## is a positive integer then our solution will be of the form,

##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##

## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##​
2. ##z=(x+iy)^\frac{1}{n}## with ##n## being a positive integer then our solution will be of the form,

##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##

## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##​
@Mark44 like you said it does not really matter whether ##n## is positive or fractional index.​
Probably the only variation would be on say;​
3. ##z^n=x^\frac{n_1}{n_2}## where ##n_1## and ##n_2## are integers then, our solution will be of the form,​
##z=r^\frac{n_1}{n_2}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{n_1}{n_2}##

## =r^\frac{n_1}{n_2}\left[\dfrac{\cos (n_1(θ+2kπ)}{n_2}+\dfrac{\sin n_1(θ+2kπ)}{n_2} \right]##​

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chwala said:
I can summarise as follows with your indulgence; if we are solving;

1. ##z^n=x^n## where ##x## is say an integer value i.e from ##z=x+iy## with ##y=0## and ##n## is a positive integer then our solution will be of the form,

##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##

## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##​
1. First off, your formula is wrong. It's not ##\frac 1 n \cos(\theta + \dots )## etc. It's ##\cos(\frac{\theta + \dots}n)## etc. See the end of my post for De Moivre's Formula.
2. If y = 0 (i.e., Im(z) = 0), then z is purely real and you don't need to go through all of this. In that case, z and ##z^n## will lie along the real axis and θ will be either 0 or ##2\pi##.
##z^n = x^n \Rightarrow z = \pm x##, with the sign depending on whether n is even or odd.
3. x can be real, which includes integers and rational numbers.
chwala said:
2. ##z=(x+iy)^\frac{1}{n}## with ##n## being a positive integer then our solution will be of the form,

##z=r^\frac{1}{n}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{1}{n}##

## =r^\frac{1}{n}\left[\dfrac{\cos (θ+2kπ)}{n}+\dfrac{\sin (θ+2kπ)}{n} \right]##​
As already noted for your first case, the above is incorrect. The angle is what gets divided/multiplied, not the cosine or sine of the angle.
chwala said:
@Mark44 like you said it does not really matter whether ##n## is positive or fractional index.​

Probably the only variation would be on say;​

3. ##z^n=x^\frac{n_1}{n_2}## where ##n_1## and ##n_2## are integers then, our solution will be of the form,​

##z=r^\frac{n_1}{n_2}\left[\cos (θ+2kπ)+ i \sin (θ+2kπ)\right]^\frac{n_1}{n_2}##

## =r^\frac{n_1}{n_2}\left[\dfrac{\cos (n_1(θ+2kπ)}{n_2}+\dfrac{\sin n_1(θ+2kπ)}{n_2} \right]##​
You don't need separate cases for #2 and #3.
If z = x + iy, where x and y are real numbers, then
##z^n = r^n [\cos(n(\theta + 2k\pi)) + i \sin(n(\theta + 2k\pi))] = r^n cis(n(\theta + 2k\pi)##
If n is a positive integer, you can omit the ##2k\pi## terms.
If n = 1/m, where m is a positive integer, k will be in the range 0, ..., m-1.
If n is a rational number in reduced form a/b, then ##z^n = z^{a/b} = (z^{1/b})^a##, so the above apply.

The basic idea when a complex number z is written in polar form is this:
##z = r cis(\theta) \Rightarrow z^n = r^n cis(n \theta)##
To raise a complex number to a power, raise the modulus (magnitude) to that power and multiply the argument (angle) by that power -- this is called De Moivre's Formula. See https://en.wikipedia.org/wiki/De_Moivre's_formula.

One more thing. Your use of INDENT tags is really cluttering things up. Please limit your use of such things.

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chwala
@chwala, one more comment sort of related to your use of INDENT tags. The formulas in this thread are relatively complicated, and easy to get wrong when crafting TeX for them, and I'm speaking from personal experience. While the effort you put into formatting your post is much appreciated, you should try to balance that effort against the cost of that presentation -- a greater chance for making an error.
Regarding the INDENT tags, I noticed a slew of open and close INDENT tags that had nothing between them, so they weren't doing anything but creating clutter. I deleted them to make it easier to insert my comments at the relevant places.

chwala

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