De Moivre's formula for complex roots

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Mondo

Junior Member
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Hello,

I have some troubles understanding why De Moivre's formula for complex roots if valid. It says, if `z = r(cos(\theta) + isin(\theta))` then `z^n = r^n(cos(n\theta) + isin(n\theta))` and we can get roots by: `r^(1/n) (cos((\theta + 2k\pi)/n) + isin((\theta + 2k\pi)/n))`, number of roots -> `k = 0, ..., n-1`.

Now they say "While it is not true that any real number has `n` real nth roots, every complex number (and thus every real number) has n complex roots. " - So what I don't get is why in the same sentence they claim not every real number has nth roots but in the same time, when treated as complex numbers they have?

Thanks
 
Mondo said:
Now they say "While it is not true that any real number has `n` real nth roots, every complex number (and thus every real number) has n complex roots. " - So what I don't get is why in the same sentence they claim not every real number has nth roots but in the same time, when treated as complex numbers they have?
Click to expand...
The operative words are real roots: there are only two real fourth roots of \(16\).
On the other hand there are four complex fourth roots of \(16\). SEE HERE
 
It is I think a very carelessly written sentence. Complex numbers and real numbers are different number systems. Every real number a has a corresponding complex number, namely a + 0i. Complex numbers have complex roots. Real numbers do not have complex roots. What they are trying to say is that a complex number corresponding to a real number may have complex roots that do not correspond to any real root of the corresponding real number.

For example,

[MATH]\text {There is no real number } a \text { such that } a^2 = - 4.[/MATH]
However

[MATH](0 + 2i)^2 = 0^2 + 2(0)(2i) + (2i)^2 = 4i^2 + 0 = -4 + 0i.[/MATH]
[MATH](0 - 2i)^2 = 0^2 - 2(0)(2i) + (-2i)^2 = 4i^2 + 0 = -4 + 0i.[/MATH]
For another example

[MATH]\text {There is exactly one real number } a \text { such that } a^3 = - 1, \text {namely } a = -1.[/MATH]
However

[MATH](- 1+ 0i)^3 = (-1)^3 + 3(-1)^2(0) + 0 + 3(-1)(0)^2 + (0i)^3 = -1 + 0i. [/MATH]
[MATH]\left ( \dfrac{1 + i\sqrt{3}}{2} \right )^3 = \dfrac{1^3 + 3(1^2)(i \sqrt{3}) + 3(1)(i \sqrt{3})^2 + i^3(\sqrt{3})^3}{8} = \dfrac{1 + 3i\sqrt{3} + (3)(-3) - 3i\sqrt{3} }{8} = - 1 + 0i.[/MATH]
[MATH]\left ( \dfrac{1 - i\sqrt{3}}{2} \right )^3 = \dfrac{1^3 - 3(1^2)(i \sqrt{3}) + 3(1)(i \sqrt{3})^2 - i^3(\sqrt{3})^3}{8} = \dfrac{1 - 3i\sqrt{3} + (3)(-3) + 3i\sqrt{3} }{8} = - 1 + 0i.[/MATH]
Now that you have seen some examples, you can see what should have been said is something like

“Ignoring the special cases of 2nd roots and zero, real numbers do not have n nth real roots, but the corresponding complex numbers do have n nth complex roots like all other non-zero complex numbers.”
 
JeffM said:
It is I think a very carelessly written sentence. Complex numbers and real numbers are different number systems. Every real number a has a corresponding complex number, namely a + 0i. Complex numbers have complex roots. Real numbers do not have complex roots. What they are trying to say is that a complex number corresponding to a real number may have complex roots that do not correspond to any real root of the corresponding real number.
Click to expand...
Why would you write that. The two square roots of \(-4\) are \(\pm2i\).
So there is an example of a real number having two complex roots.
There are four fourth roots of \(-16\) SEE HERE, each of which is complex.
Consider the eight eighth roots of \(256\) SEE HERE, a mixture of real & complex.
 
pka said:
Why would you write that. The two square roots of \(-4\) are \(\pm2i\).
So there is an example of a real number having two complex roots.
There are four fourth roots of \(-16\) SEE HERE, each of which is complex.
Consider the eight eighth roots of \(256\) SEE HERE, a mixture of real & complex.
Click to expand...
I wrote that because complex numbers are defined as being in the form a + bi, where a and b are real numbers. And - 4 is not in the form of a complex number. It is lazy and sloppy to say that rational numbers are integers or that integers are whole numbers. It is equally lazy and sloppy to say that complex numbers are real numbers.
 
JeffM said:
I wrote that because complex numbers are defined as being in the form a + bi, where a and b are real numbers. And - 4 is not in the form of a complex number. It is lazy and sloppy to say that rational numbers are integers or that integers are whole numbers. It is equally lazy and sloppy to say that complex numbers are real numbers.
Click to expand...
But that is not what you said! You did not say that -4 was a real number (which is true) you said that "real numbers do not have complex roots" which is false.

Perhaps you were arguing that the roots of a real number are either real or imaginary and you are taking "complex number" to mean "a+ bi" and requiring that neither a nor b to be 0. If so that is certainly NOT the standard definition of "complex number".

You say "It is lazy and sloppy to say that rational numbers are integers or that integers are whole numbers. It is equally lazy and sloppy to say that complex numbers are real numbers."

But we are not saying either of those things! We (or at least I) would say that "integers are rational numbers and that whole numbers are integers", the opposite of what you have. We are here saying that "real numbers are complex numbers", NOT that "complex numbers are real numbers.

The standard definition of "complex number" is a number or the form a+ bi where a and b can be any real numbers, including 0. The set of real numbers, numbers of the form a+ 0i, is a subset of the complex numbers, as is the set of pure imaginary numbers, of the form 0+ bi.
 
@HallsofIvy

According to you, the standard definition “of a complex number is a number of the form a + bi.“ Is - 4 in that form? No. We are in the realm of real numbers. If we are talking about a complex number such as - 4 + 0i, we should call it a complex number

The natural numbers are not closed under subtraction. That is a fact about natural numbers. When we are talking about natural numbers, an expression like 4 - 6 is meaningless. When we talk about the one point compactification of the real numbers, a / 0 is a number if a is not zero. Being clear about the realm of discourse facilitates communication.

Perhaps it is being pedantic to insist on being careful about terminology or notation. But it avoids confusion. 1 has one cube root (a true statement if we are talking about real numbers). 1 + 0i has three cube roots. If we view 1 + 0i as unnecessarily complex, at least say the complex number 1 has three cube roots. - 4 has no square root (a true statement if we are talking about real numbers). - 4 + 0i has two distinct square roots. If we are talking about - 4 as a member of the set of complex numbers and cannot be bothered to make that clear using the standard notation of - 4 + 0i, then it avoids mystification to say the complex number - 4 has two distinct roots. Why avoid the correct notation and a correct usage in favor of an abbreviated notation and a usage that is correct but with misleading connotations about the realm of discourse.

In short, consistency matters. If we are talking about complex numbers, I am happy to say -4 is a complex number although it is not presented in standard form. And obviously it is clear what we are talking about if we say -4 + 0i. But if we ARE talking about complex numbers, then why imply that -4 is not a complex number by calling it a real number. “Exprssio unius est exclusio alterius” describes how people actually interpret statements, not just lawyers.
 
Thanks for the discussion, I think we are on the same page now and all is clear
 
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