Question

This might be stupid, but how do I factor \(z^4+1\).

I can do it by calculating the roots the long way and then write, it out, but what are the algebraic steps that get me from

\(z^4+1\) to\((z+i)(z-i)(z+1)(z-1)\)

Answer 1

First off, those aren't the actual roots of \(z^4+1\). Second, you could notice that \(z^4+1\) is the difference of two squares.\[z^4+1=z^4-(-1)=(z^2-i)(z^2+i)\]Each of these factors is then also a difference of two squares.

Answer 2

right, I just realized that haha.

Answer 3

why are there 4 King George pictures?

Answer 4

#dontquestionitgowithit

Answer 5

King Friday.. Daniel Striped Tiger..Mr. Rogers Neighborhood

Answer 6

I love KingGeorge











To make this post relevant I will say that math is awesome and I have no idea how to solve this question but good luck ;)

Answer 7

You can put it in binomials then separate it into a trinomial.

Answer 8

The second step is then\[(z^2-i)(z^2+i)=(z+\sqrt{i})(z-\sqrt{i})(z+\sqrt{-i})(z-\sqrt{-i}).\]If you were to write out \(\sqrt{i}\) and \(\sqrt{-i}\) in their exponential forms, you would get a possibly nicer looking factorization.

Answer 9

There are other ways, but that's a quick method.

Answer 10

yes thank you! I was looking for the quick method. The way I did it was to write \[z^4=e^{i(\pi/2+2\pi n)}\]and work from there, but I was just curious. Thank you.

Answer 11

oops!

Answer 12

\(e^{i( \pi+2\pi n)}\)

Answer 13

That's another way you could do it. Then\[z=e^{i(\pi+2\pi n)/4}=e^{i\pi/4+i\pi n/2}.\]Every time \(n\) increases by 4, you get the same value. So the roots will be \[e^{i\pi/4+i\pi/2},e^{i\pi/4+i\pi},e^{i\pi/4+i\pi 3/2},e^{i\pi/4}\]

Answer 14

Put more simply, those are just\[e^{i\pi/4},e^{3i\pi/4},e^{5i\pi/4},e^{7i\pi/4}.\]