Question

Questions about roots of complex numbers.

Answer 1

What's your question? :)
Do you understand the factor method?

Answer 2

i do not.

Answer 3

Recall your difference of squares formula for real numbers:\[\large\rm a^2-b^2=(a+b)(a-b)\]

Answer 4

When you have the difference of squares,
they factor into the `product of conjugates` as shown above.

Answer 5

So what they did is this:\[\large\rm z^4=1\]Subtracting 1 from each side,\[\large\rm z^4-1=0\]Rewriting it like this:\[\large\rm (z^2)^2-(1^2)^2=0\]k?

Answer 6

Which, by our formula, factors into the product of conjugates:\[\large\rm (z^2+1^2)(z^2-1^2)=0\]

Answer 7

ohhh okay

Answer 8

In this second set of brackets, we can apply the rule again.\[\large\rm (z^2+1^2)\color{orangered}{(z^2-1^2)}=0\]Do you see how?

Answer 9

yes i do

Answer 10

\[\large\rm (z^2+1^2)\color{orangered}{(z+1)(z-1)}=0\]k cool

Answer 11

The sum of squares is a little different.
There is no way to factor that into real factors.
it instead factors into `the product of complex conjugates`.

Answer 12

Example:\[\large\rm (x+iy)(x-iy)\quad=\quad x^2-ixy+ixy-(iy)^2\]\[\large\rm =x^2-(iy)^2\quad=\quad x^2-(-y^2)\quad=\quad x^2+y^2\]So notice that when you multiply `complex` conjugates, you get a sum instead of a difference.
Just slightly different than what we were doing before :)

Answer 13

\[\large\rm (z^2+1^2)(z+1)(z-1)=0\]So we'll get,\[\large\rm (z+1i)(z-1i)(z+1)(z-1)=0\]

Answer 14

yes i see

Answer 15

And you're comfortable with your `Zero-Factor Property` at this point?
Setting each factor equal to zero individually and solving for z, yes

Answer 16

yes

Answer 17

Good :)
So the factoring is not too bad.
Just gotta remember some cool tricks.

Answer 18

`nth root method`.
this method is a lot more fun in my opinion.
maybe a little confusing at first though

Answer 19

So if we're over here at this point 1+0i