Solve for z, z^4 - i = 0?

Question

Find z, and i is the sqrt(-1).

Thanks

Answer 1

We see that:



z^4 - i = 0

==> z^4 = i

==> z = i^(1/4).



So this is equivalent to trying to find the 4 fourth roots of i.



In polar form, we see that:



i = cos(π/2) + i sin(π/2).



By De Movire's Theorem:



i^(1/4) = cos[(π/2 + 2πk)/4] + i sin[(π/2 + 2πk)/4]

==> i^(1/4) = cos(π/8 + πk/2) + i sin(π/8 + πk/2) for k = 0, 1, 2, and 3.



Thus:



k = 0 ==> z = cos(π/8) + i sin(π/8) ≈ 0.9239 + 0.3827i

k = 1 ==> z = cos(5π/8) + i sin(5π/8) ≈ -0.3827 + 0.9239i

k = 2 ==> z = cos(9π/8) + i sin(9π/8) ≈ -0.9239 - 0.3827i

k = 3 ==> z = cos(13π/8) + i sin(13π/8) ≈ 0.3827 - 0.9239i.



I hope this helps!

Answer 2

z x 11 1 1 1 1 1 1 13 3x ^i. I dont know 3?

Answer 3

first subtract the i and you get z^4=i and the do the fourth root of each side to get z=the fourth root of i which is no real answers

Answer 4

z^4-i = 0



Oduzmi -i s obe strane:

z^4 = i



Neka je z = r (cos(alfa) + i sin(alfa)) i zapisi i u "fazor" (ne znam sta je to) formi:

r^4 (cos(theta)+i sin(theta))^4 = cos(pi/2)+i sin(pi/2)



Iskoristi Moavrovu teoremu:

r^4 (cos(4 alfa)+i sin(4 alfa)) = cos(pi/2)+i sin(pi/2)



Izjednacavanjem modula i argumenata dobijamo:

r^4 = 1, i ovo --> 4 alfa = 2 pi k+pi/2



Resavanjem ovih jednacina i odabirom 4 zasebna resenja, dobijamo:

r = 1 and (alfa_1 = pi/8 ili alfa_2 = (5 pi)/8 ili alfa_3 = (9 pi)/8 ili alfa_4 = (13 pi)/8)

Onda jednacina z^4 = i ima resenja:

z = cos(pi/8)+i sin(pi/8) ili z = i cos(pi/8)-sin(pi/8) ili z = -cos(pi/8)-i sin(pi/8) ili z = -i cos(pi/8)+sin(pi/8)