roots- Complex Numbers

[Syndicate]

Hey

I was recently attempted to solve this question: Find the forth roots of 16i, however,  failed to do so.

Any assistance appreciated 

[Callum@1373]

These questions are beyond 1&2 spesh I believe so don't try to do anything you aren't ready for  but I answered this question in the specialist question thread

[Syndicate]

These questions are beyond 1&2 spesh I believe so don't try to do anything you aren't ready for  but I answered this question in the specialist question thread

My idea was to find 4 solutions for 16i, but just wanted to make sure if I am at the right track

[Callum@1373]

My idea was to find 4 solutions for 6i, but just wanted to make sure if I am at the right track

I think it was 16i not 6i, not sure if you understand but it would be more preferable because the radius of the four roots of 16i will be 2 but the radius for the four roots of 6i will be 6 to the power of 1/4, which is nasty

[Syndicate]

I think it was 16i not 6i, not sure if you understand but it would be more preferable because the radius of the four roots of 16i will be 2 but the radius for the four roots of 6i will be 6 to the power of 1/4, which is nasty

yeah, its 16i. Sorry about the typo 

[cosine]

To find the fourth roots of 16i, we are basically finding the fourth root of z.

Say z = 16i, now to find the fourth roots, we need to fourth root (not square root, not cube root, but fourth root) z, and you can do this using De Moivre's Theorem, stating that:







So our first step is to convert z=16i into polar form, and you should get

So now we can apple De Moivre's Theorem: 

So the final answer would be:

You now need to add 2pi/n, where n is the number of solutions required. You do this because it asks for the fourth roots, and hence there are 4 solutions to this problem.

Hope this helps guys

[brightsky]

To find the fourth roots of 16i, we are basically finding the fourth root of z.

Say z = 16i, now to find the fourth roots, we need to fourth root (not square root, not cube root, but fourth root) z, and you can do this using De Moivre's Theorem, stating that:







So our first step is to convert z=16i into polar form, and you should get

So now we can apple De Moivre's Theorem: 

So the final answer would be:

@callum, the reason why you don't add 2pi/n to each solution is because you're not solving , you're simply taking the fourth ROOT of the complex number, which will yield only one answer!

Hope this helps guys

This, except just to clarify that 16i has 4 fourth roots. The definition of the fourth root of 16i is the root/solution of the equation z^4 = 16i.

[Callum@1373]

To find the fourth roots of 16i, we are basically finding the fourth root of z.

Say z = 16i, now to find the fourth roots, we need to fourth root (not square root, not cube root, but fourth root) z, and you can do this using De Moivre's Theorem, stating that:







So our first step is to convert z=16i into polar form, and you should get

So now we can apple De Moivre's Theorem: 

So the final answer would be:

@callum, the reason why you don't add 2pi/n to each solution is because you're not solving , you're simply taking the fourth ROOT of the complex number, which will yield only one answer!

Hope this helps guys

That's not quite right, similar to brightsky said, for a polynomial of n degree their are n roots

Sydnicate,refer to my answer in the question thread

[lzxnl]

If a question says 'the fourth root', you actually then only take one of the four solutions to z^4=16i
That is called the principal root

But this question has 'roots', plural, so you are supposed to find all of the solutions

[cosine]

That's not quite right, similar to brightsky said, for a polynomial of n degree their are n roots

Sydnicate,refer to my answer in the question thread

Actually, no. It depends on the wording of the questions. Besides, the rule "for a polynomial of n degree their are n roots" does not hold for all values in the interval , so then would have 0.25 solutions?

As lzxnl said above, depends on whether its plural or not, in this case it does say roots and not root, which was a misconception as Syndicate's post on the Specialist thread stated root, and not roots.

[Callum@1373]

Actually, no. It depends on the wording of the questions. Besides, the rule "for a polynomial of n degree their are n roots" does not hold for all values in the interval , so then would have 0.25 solutions?

As lzxnl said above, depends on whether its plural or not, in this case it does say roots and not root, which was a misconception as Syndicate's post on the Specialist thread stated root, and not roots.

That rule is for only positive integers i believe. But yeah, gotta have roots to get multiple solutions

Edit: That last sentence could be viewed in a really, really dirty way ahahahhaha

[lzxnl]

That rule is for only positive integers i believe. But yeah, gotta have roots to get multiple solutions

Edit: That last sentence could be viewed in a really, really dirty way ahahahhaha

Everything can be dirty if you try hard enough.

That rule is in general so that you can define a function f(z) = z^n over the complex numbers for any n real or complex

[abeybaby]

Everything can be dirty if you try hard enough.

Hehe, you said hard...

[brightsky]

If a question says 'the fourth root', you actually then only take one of the four solutions to z^4=16i
That is called the principal root

But this question has 'roots', plural, so you are supposed to find all of the solutions

Actually, no. It depends on the wording of the questions. Besides, the rule "for a polynomial of n degree their are n roots" does not hold for all values in the interval , so then would have 0.25 solutions?

As lzxnl said above, depends on whether its plural or not, in this case it does say roots and not root, which was a misconception as Syndicate's post on the Specialist thread stated root, and not roots.

The following article should help to clear things up: http://mathworld.wolfram.com/nthRoot.html. In particular, I refer to the following passages:

"The nth root of a quantity z is a value r such that z = r^n."

"Rolle proved that any complex number has exactly n nth roots (Boyer 1968, p. 476), though some are possibly degenerate. However, since complex numbers have two square roots and three cube roots, care is needed in determining which root is under consideration. For complex numbers z, the root of interest (generally taken as the root having smallest positive complex argument) is known as the principal root. However, for real numbers, the root of interest is usually the root that is real (when it exists)."

As lzxnl mentioned above, if a question asks for the root of a particular number, then it is likely that it is asking for the principle square root, which, as stated above, is generally taken as the root with the smallest positive complex argument. For example, if a question asks for the square root of 4, it is likely that it is asking for the principle square root of 4, which is 2. But there are in actual fact two square roots of 4, namely 2 and -2.

One formulation of the Fundamental Theorem of Algebra is that a polynomial of degree n has exactly n complex roots, some of which may be repeated. A polynomial in one variable is defined as any expression that can be written in the form a_n x^n + ... + a_2 x^2 + a_1 x + a_0, where a_n, ..., a_2, a_1 and a_0 are all constants and n, ..., 2, 1, 0 are all non-negative integers. Hence, by the very definition of a polynomial, n has to be a non-negative integer. z^(1/4) is not a polynomial, and so the Fundamental Theorem of Algebra does not hold. But as mentioned above, z^(1/4), which in itself is quite an ambiguous way to represent the fourth root of z, is defined to be a value r such that z = r^4. This is a polynomial equation of degree 4, and so has 4 complex solutions, which are said to be the fourth roots of z.

Hope this clarifies any confusion!