Question

HOW TO find the imaginary solution for x^4+1=0??
pLEASE HELP....

Answer 1

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

Answer 2

it is not factorable , I thought

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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}

Answer 11

please see my answer above ... those are roots of 1.

Answer 12

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

Answer 13

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.