why are there imaginary solutions to equation "x^4 = 256"?

[allegansveritatem]

Problem:

. . . . .49 [math]x^4\, =\, 256[/math]
What I did:

. . . . .[math]x^4\, =\, 256[/math]
. . . . .[math]x\, =\, \pm \sqrt[4]{256\,}[/math]
. . . . .[math]x\, =\, \pm 4 \qquad \mbox{(my solution)}[/math]
. . . . .[math]x\, =\, \pm 4,\, \pm 4i \qquad \mbox{(book's solution)}[/math]
What i want to know is: Where did they get that plus/munus 4i? I know that 4i to the 4th power is 256 because I used a calculator to test it but I don't know why or how this is come by.

 

[Dr.Peterson]

Problem:

. . . . .49 [math]x^4\, =\, 256[/math]
What I did:

. . . . .[math]x^4\, =\, 256[/math]
. . . . .[math]x\, =\, \pm \sqrt[4]{256\,}[/math]
. . . . .[math]x\, =\, \pm 4 \qquad \mbox{(my solution)}[/math]
. . . . .[math]x\, =\, \pm 4,\, \pm 4i \qquad \mbox{(book's solution)}[/math]
What i want to know is: Where did they get that plus/munus 4i? I know that 4i to the 4th power is 256 because I used a calculator to test it but I don't know why or how this is come by.

Every nonzero number has four fourth roots (in the complex numbers). Presumably the context here assumes complex numbers, and you have been taught how to find the n nth roots.

Look back in the book to find this information. I'd tell you what specifically to look for, but there are a couple different things they might have taught, and I don't want to confuse you by talking about something different. If you tell us the title of the section, and the objectives it covers (or list the theorems), that will probably give us a better idea of what you are expected to do.

 

[Otis]

… Where did they get that plus/minus 4i? … I used a calculator to test it …

I can't tell whether you're asking about the ± sign, or why i^4 equals one, or both. ?

Are you familiar with the periodic behavior of i^n, for increasing Natural numbers n ?

?

 

[pka]

Problem:

. . . . .49 [math]x^4\, =\, 256[/math]
What I did:

. . . . .[math]x^4\, =\, 256[/math]
. . . . .[math]x\, =\, \pm \sqrt[4]{256\,}[/math]
. . . . .[math]x\, =\, \pm 4 \qquad \mbox{(my solution)}[/math]
. . . . .[math]x\, =\, \pm 4,\, \pm 4i \qquad \mbox{(book's solution)}[/math]
What i want to know is: Where did they get that plus/munus 4i? I know that 4i to the 4th power is 256 because I used a calculator to test it but I don't know why or how this is come by.

Factor \(\displaystyle 256=2^8\) and the root is \(\displaystyle \zeta=2^2\exp(0\cdot\bf{i})\).
Every number has four fourth roots all of which lie on a circle with center at \(\displaystyle (0,0)\) and radius \(\displaystyle 4\)( in this case because \(\displaystyle (256)^{1/4}=4)\)
Now the rotations are achieved by \(\displaystyle \rho=\exp\left(\frac{2\pi \bf{i}}{4}\right)\)
Now this setup makes the roots a snap, they are: \(\displaystyle \zeta\cdot\rho^k,\;k=0,1,2,3\)

 

[MarkFL]

Yes, there are many ways to approach this, perhaps the most basic is by factoring:

[MATH]x^4=256[/MATH]
[MATH]x^4-256=0[/MATH]
[MATH]x^4-4^4=0[/MATH]
[MATH](x^2-4^2)(x^2+4^2)=0[/MATH]
Now, since \(i^2=-1\) we can write:

[MATH](x^2-4^2)(x^2-(4i)^2)=0[/MATH]
[MATH](x+4)(x-4)(x+4i)(x-4i)=0[/MATH]
Hence:

[MATH]x\in\{\pm4,\pm4i\}[/MATH]

 

[allegansveritatem]

Every nonzero number has four fourth roots (in the complex numbers). Presumably the context here assumes complex numbers, and you have been taught how to find the n nth roots.

Look back in the book to find this information. I'd tell you what specifically to look for, but there are a couple different things they might have taught, and I don't want to confuse you by talking about something different. If you tell us the title of the section, and the objectives it covers (or list the theorems), that will probably give us a better idea of what you are expected to do.

I have not run across quartic equations--except some that are factored into quadratics. I have decided not to skip the first 60 pages of the algebra review of my new book. There is lot in there that I haven't been introduced to. The section this problem comes from deals with complex numbers and how to operate with them.

 

[allegansveritatem]

I can't tell whether you're asking about the ± sign, or why i^4 equals one, or both. ?

Are you familiar with the periodic behavior of i^n, for increasing Natural numbers n ?

?

yes, I know something about complex numbers and the values of i and i squared, cubed and brought to 4th power.

 

[allegansveritatem]

Factor \(\displaystyle 256=2^8\) and the root is \(\displaystyle \zeta=2^2\exp(0\cdot\bf{i})\).
Every number has four fourth roots all of which lie on a circle with center at \(\displaystyle (0,0)\) and radius \(\displaystyle 4\)( in this case because \(\displaystyle (256)^{1/4}=4)\)
Now the rotations are achieved by \(\displaystyle \rho=\exp\left(\frac{2\pi \bf{i}}{4}\right)\)
Now this setup makes the roots a snap, they are: \(\displaystyle \zeta\cdot\rho^k,\;k=0,1,2,3\)

I don't know trigonometry which I suppose these expressions come from.

 

[allegansveritatem]

Yes, there are many ways to approach this, perhaps the most basic is by factoring:

[MATH]x^4=256[/MATH]
[MATH]x^4-256=0[/MATH]
[MATH]x^4-4^4=0[/MATH]
[MATH](x^2-4^2)(x^2+4^2)=0[/MATH]
Now, since \(i^2=-1\) we can write:

[MATH](x^2-4^2)(x^2-(4i)^2)=0[/MATH]
[MATH](x+4)(x-4)(x+4i)(x-4i)=0[/MATH]
Hence:

[MATH]x\in\{\pm4,\pm4i\}[/MATH]

I followed you until you suddenly introduced i^2. There is a skipped or compressed step in here somewhere where the radicals with the negative radicands come in. No?

 

[MarkFL]

I followed you until you suddenly introduced i^2. There is a skipped or compressed step in here somewhere where the radicals with the negative radicands come in. No?

I introduced \(i^2\) to allow the sum of two squares to be written as the difference of two squares, which can then be factored.

 

[allegansveritatem]

Yes, there are many ways to approach this, perhaps the most basic is by factoring:

[MATH]x^4=256[/MATH]
[MATH]x^4-256=0[/MATH]
[MATH]x^4-4^4=0[/MATH]
[MATH](x^2-4^2)(x^2+4^2)=0[/MATH]
Now, since \(i^2=-1\) we can write:

[MATH](x^2-4^2)(x^2-(4i)^2)=0[/MATH]
[MATH](x+4)(x-4)(x+4i)(x-4i)=0[/MATH]
Hence:

[MATH]x\in\{\pm4,\pm4i\}[/MATH]

I will have to look at this closely tomorrow--my brain doesn't work after 11--and try to explain it to myself. I will get back. Thanks for working it out.

 

[allegansveritatem]

well, while copying this out I think I got what you did with the i^2--it was used to allow the factoring of the sum of two squares which in the sum form is not factorable. Good. I get it. No?