1. anonymous

To get it I would factor it into (x^2+1)(x^2-1) [sum of perfect squares] and then factor these to (x+1)(x-1)(x+i)(x-i) so your imaginary solutions are i and -i

2. anonymous

it is not factorable , I thought

3. anonymous

oops I just realized my first factors are in fact incorrect. I misread the original as x^4-1

4. anonymous

x^4=-1 x=fourth root of -1

5. anonymous

the answer is 1+i/square root of 2,, but I dont know why

Since equation is of order 4, we get four different solutions in complex domain. 1+0i -1+0i 0+1i 0-1i or simply $\pm i\mbox{ , }\pm 1$

7. anonymous

square root of -1 is i square root of i is -i and i

8. anonymous

Instead factor it to (x^2 + i)(x^2 - i)

9. anonymous

yes.. I did, but ten what's square root of I ??

10. anonymous

So from the quadratic formula we get (1+√i)/2, (1-√i)/2, (-1+√i)/2, and (-1-√i)/2 but as for √i..... I suppose you could rewrite it as (-1)^{1/4}

12. anonymous

I agree that $\pm{i}$ are solutions but 1 and -1 are not. If you go back to the original problem and substitute them in you get an incorrect statement. 1^4 + 1≠0

I am sorry ... I have done a terrible mistake in the previous calculation. given equation $x^4=-1$ since $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ So, we can write $-1=e^{i\pi}$ So, we can say $x=(-1)^{1/4}=e^{\frac{i\pi}{4}}=\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}$ $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ Similarly taking diffrent values of $\theta$ we can get $\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$ $-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ $-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$ are also roots for the given equation.