5th roots of a complex number

IsThisIt???

Can anyone tell me what the 5th roots of the complex number z=-1-i are.
I've tried doing it out but get crazy answers that I don't think could be right.

I think my mistake might be the angle I'm using, 225deg. I got this from finding arctan-1/-1=45deg and as seen as it is in the 3rd quad I added 180deg. Is this wrong?

Any help is much appreciated

Michael Collins

That angle seems correct. Are you familiar with the polar representation of complex numbers and Euler's formula?

Euler's formula says:

[latex] \displaystyle r e^{i\theta} = r(\cos(\theta) + i\sin(\theta))[/latex]

So this relates the Cartesian, i.e. real and imaginary form of a complex number, to the polar, i.e. magnitude and angle form. This latter form makes it very easy to take roots.

So your number in Cartesian format is

[latex] \displaystyle -1 - i[/latex]

which has angle [latex] 225^\circ [/latex] as you correctly calculated (we could also express this as [latex] -135^\circ [/latex], same thing) and magnitude

[latex] \displaystyle \sqrt{\Re^2 + \Im^2}= \sqrt2 = 2^\frac{1}{2}[/latex]

so in polar form this number is:

[latex] \displaystyle z = 2^\frac{1}{2}e^{i225^\circ}[/latex]

but if we add [latex] 360^\circ [/latex] to this we still have the same number so really:

[latex] \displaystyle z = 2^\frac{1}{2}e^{i(225^\circ + n360^\circ)}[/latex]

Now you want the fifth roots, so [latex]z^{1/5}[/latex], which gives

[latex] \displaystyle z^\frac{1}{5} = (2^\frac{1}{2})^\frac{1}{5} (e^{i(225^\circ + n360^\circ)})^\frac{1}{5}[/latex]

this will give five distinct answers for different integer values of n. I'll do the first one and you can have a go at the others

[latex] \displaystyle z^\frac{1}{5} = (2^\frac{1}{10})(e^{i(45^\circ + n72^\circ)}) = 2^\frac{1}{10} e^{i45^\circ} \hbox{for } n=0[/latex] since exponents are multiplied when combined like above.

Do this yourself for 4 more values of n, and you should get 4 more answers, all different. Do it for a fifth and you'll repeat one of the answers you've already got.

Also you may have to convert back to Cartesian form if this is for an exam. So for my example, you'd have

[latex] \displaystyle z^\frac{1}{5} = 2^\frac{1}{10} e^{i45^\circ} = 2^\frac{1}{10} (1 + i) = 2^\frac{1}{10} + i2^\frac{1}{10}[/latex]

but really the polar (i.e. exponential form) is just as valid a way to represent a complex number as any.

IsThisIt???

Okay I follow everything about the exponential form and think I've got it up to there.
Just one question about converting it into cartesian form. Why doesn't it include 1/(sqrt)2 since sin45 and cos45 equal 1/(sqrt)2? When I do it now I get (2^(1/10))(1/sqrt2 + i1/sqrt2) for n=0

Michael Collins

Ack, you're dead right. For some reason I forgot to include that when converting from the polar form of [latex] e^{i45^\circ} [/latex] to the Cartesian, which is of course

[latex] \displaystyle \frac{1}{\sqrt{2}}(1+i). [/latex]

My mistake - well spotted!

IsThisIt???

Got the other ones out too. Thanks for your help

Fremen

Just to be pedantic here - you really shouldn't write exp(45 degrees). It doesn't make sense to exponentiate a degree, since a degree has dimension.

Think about the definition of exp. If you express the same angle in grads, compute the exponential, then change back to units of degrees, will you get the same answer as if you had done the whole calculation in degrees? I haven't checked this, but I'm guessing you won't. Does it even make sense to square a degree?

Michael Collins

Yeh that's a valid point alright, and I've rarely seen degrees used like above to be fair. I just thought it was easier from a pedagogical point of view.

Of course, you could argue that degrees are really unitless quantities, much like radians - and no one has any qualms exponentiating radians.

Michael Collins

I've been thinking about this some more. It's actually not that pedantic at all really, quite worth pointing out. Although I don't think IsThisIt??? should worry about it too much just yet - keep it in mind though.

If you define the exponential as the standard power series, as you say, just plugging in the degrees won't work out (I haven't done this either, but I don't see how it can possibly work), and I don't see how squaring (or any higher power of) a degree measure makes any sense. You basically have to say the degrees notation is a short hand - whereby you ultimately substitude radians when such calculations are required (I assume you meant radians above, as I reckon gradians have the same issues).

It interesting how in physical forumlae that are derived from basic physical principles, any exponents are generally unitless - in fact it is often a good check to see how physically valid a formula is. In electronic circuits, for example, you can show that if you charge a capacitor up to a certain voltage [latex]E[/latex], then discharge it through a resistor, the discharging formula for the voltage across the capacitor is

[latex] \displaystyle v(t) = Ee^{-\frac{t}{RC}} [/latex]

[latex]t[/latex] here being time. For the exponent to be unitless, it means that the product of the resistance and the capacitance must be in units of time - which we can show it is! Not something that's immediately obvious though I would have thought.

MathsManiac

It might be worth noting that, even though the identity [latex] \displaystyle r e^{i\theta} = r(\cos(\theta) + i\sin(\theta))[/latex] isn't really true when theta is in degrees, it is nonetheless still the case that [latex] \displaystyle {(r(\cos\theta + i\sin\theta))}^n = r^n(\cos(n\theta) + i\sin(n\theta))}[/latex].

On the Leaving Cert course, this is referred to as De Moivre's theorem, and is done without reference to the exponential function. Students prove it for n a natural number by induction, and extend it to integers. They accept it without proof for rational n, and apply it to problems like this one, (although a proof is not too difficult for rational n once you have it for integer n).

So, at Leaving Cert, students can do this kind of problem either in degrees or radians, without running into a dodgy use of the exponential function.

Michael Collins

I always wondered why they didn't just go one step further at Leaving Cert level and introduce Euler's formula, it makes de Moivre's theorem obvious!

As far as I can remember it is just proved for natural numbers on the LC these days (non Project Maths syllabus anyway) with negative integers stated to be also valid, and rationals stated to be true when using the general form - but no proof required for either.

It was quite a relevation to me when I went to college and figured out that what I was really doing all those times when using de Moivre's theorem was raising something to a power, I felt it should dawned on me...sort of.

I wonder if an LC class was given a hint would they make a good guess that they're really manipulating an exponential, if asked say:

Given that due to de Moivre's Theorem where [latex] f(\theta) = \cos(\theta) + i \sin(\theta) [/latex] we have

[latex] \displaystyle f(\theta)^n = f(n\cdot \theta) [/latex]

suggest another mathematical form with this property...

Not going to get the fact its the exponential function, but maybe a power of some sort? Maybe not...

Fremen

Michael Collins wrote: »

[latex] \displaystyle f(\theta)^n = f(n\cdot \theta) [/latex]

Pretty sure you can use that as a defining property for the exponential, up to choice of base.

Michael Collins

Fremen wrote: »

Pretty sure you can use that as a defining property for the exponential, up to choice of base.

Yeh exactly, some sort of power function looks likely but the exponential function [latex] g(x) = e^x [/latex] would need more constraints. I bet the property of [latex]f[/latex] from the previous post along with the following could be enough

[latex] f'(\theta) = if(\theta) [/latex]

Fremen

Michael Collins wrote: »

Yeh exactly, some sort of power function looks likely but the exponential function [latex] g(x) = e^x [/latex] would need more constraints.

My gut feeling is that there's something interesting there. You should be able to characterise [latex] e^{i \theta} [/latex] by [latex] \displaystyle f(\theta)^n = f(n\cdot \theta) [/latex],
[latex] \displaystyle f(2 \pi + \theta) = f(\theta) [/latex] and by specifying its value at one point that isn't a fraction of pi.

This gives a nice definition of the exponential without mucking about with calculus or infinite limits. I'm guessing you could take this idea pretty far, but I don't immediately see a way to extend beyond the unit circle.

RoundTower

Fremen wrote: »

Just to be pedantic here - you really shouldn't write exp(45 degrees). It doesn't make sense to exponentiate a degree, since a degree has dimension.

I don't see anything wrong with this at all. The "degree" symbol is simply shorthand for the transcendental number pi / 180. All the formulae above work just fine.

Fremen

RoundTower wrote: »

I don't see anything wrong with this at all. The "degree" symbol is simply shorthand for the transcendental number pi / 180. All the formulae above work just fine.

If that's your definition of degree then everything works fine, but what you're implicitly doing there is converting to radians. The naive interpretation would be exp(one degree) = e, which is what I was objecting to.

I'm pretty sure degrees were invented before radians, and were just thought of as an arc of a circle.

RoundTower

You're not "converting" to anything. By the same logic, the word "radians" is shorthand for "multiply by one", i.e. the identity function.

If I do a U-turn in a car, I'm turning through an angle of pi. To say I am turning through pi radians is redundant. You might as well say the car weighs 1000 kilogram-radians and it is 200 kilometer-radians from Dublin to Cork. How many people can fit in my car? Five radians, perhaps six radians if they are small.

It's not a "naive interpretation" to think that exp(one degree) should equal e, it's just wrong. As wrong as it is to think 7 + 6 equals 76, or 3/4 + 2/3 = 5/7, which are both understandable mistakes, but wrong.