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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To get it I would factor it into (x^2+1)(x^21) [sum of perfect squares] and then factor these to (x+1)(x1)(x+i)(xi) so your imaginary solutions are i and i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is not factorable , I thought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops I just realized my first factors are in fact incorrect. I misread the original as x^41

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^4=1 x=fourth root of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is 1+i/square root of 2,, but I dont know why

sgadi
 5 years ago
Best ResponseYou've already chosen the best response.0Since equation is of order 4, we get four different solutions in complex domain. 1+0i 1+0i 0+1i 01i or simply \[\pm i\mbox{ , }\pm 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0square root of 1 is i square root of i is i and i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Instead factor it to (x^2 + i)(x^2  i)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes.. I did, but ten what's square root of I ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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}

sgadi
 5 years ago
Best ResponseYou've already chosen the best response.0please see my answer above ... those are roots of 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

sgadi
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.
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