Find all solutions of z4 = -8i.
Recall:
z = a + bi, the complex form
z = |z|(cos θ + i sin θ), the polar coordinate form
So we can write:
z4 = [|z|(cos θ + i sin θ)]4 = |z|4(cos 4θ + i sin 4θ); -8i = 8(0 - i), and we have
|z|4(cos 4θ + i sin 4θ) = 8(0 - i).
Thus, |z|4 = 8, so |z|= 81/4.
The angle θ for z must satisfy cos 4θ = 0 and sin 4θ = -1.
Consequently, 4θ = 3pi/2 + 2npi for an integer n, so that θ = 3pi/8 + npi/2.
The different values of θ obtained where 0 ≤ θ ≤ 2pi are:
n = 0, θ = 3pi/8 (1st quadrant)
n = 1, θ = 3pi/8 + pi/2 = 7pi/8 (2nd quadrant)
n = 2, θ = 3pi/8 + pi = 11pi/8 ( 3rd quadrant)
n = 3, θ = 3pi/8 + 3pi/2 = 15pi/8 (4th quadrant))
Thus the solutions of z4 = -8i are
(81/4)(cos 3pi/8 + i sin 3pi8) ≈ 0.6435942529 + 1.553773974 i
(81/4)(cos 7pi/8 + i sin 7pi/8) ≈ -1.553773974 + 0.6435942529 i
(81/4)(cos 11pi/8 + i sin 11pi/8) ≈ -0.6435942529 - 1.553773974 i
(81/4)(cos 15pi/8 + i sin 15pi/8) ≈ 1.553773974 - 0.6435942529 i
8i and -8i both satisfy this: (8i)² = (8²)(i²) = (64)(-1) = -64, and (-8i)² = (-8²)(i²) = (64)(-1) = -64
8i
The square root of 64 is +/- 8
352
The square root of -64 is 8i. ' i ' is the unit imaginary number, equal to the square root of -1, or 1 at an angle of pi/2.
8i and -8i both satisfy this: (8i)² = (8²)(i²) = (64)(-1) = -64, and (-8i)² = (-8²)(i²) = (64)(-1) = -64
If you mean 8i, i might be any variable, but it may also stand for the imaginary unit, sometimes defined as the square root of minus 1. In that case, 8i is 8 times the square root of minus 1.If you mean 8i, i might be any variable, but it may also stand for the imaginary unit, sometimes defined as the square root of minus 1. In that case, 8i is 8 times the square root of minus 1.If you mean 8i, i might be any variable, but it may also stand for the imaginary unit, sometimes defined as the square root of minus 1. In that case, 8i is 8 times the square root of minus 1.If you mean 8i, i might be any variable, but it may also stand for the imaginary unit, sometimes defined as the square root of minus 1. In that case, 8i is 8 times the square root of minus 1.
8i
The square root of 64 is +/- 8
352
The square root of -64 is 8i. ' i ' is the unit imaginary number, equal to the square root of -1, or 1 at an angle of pi/2.
8i
8 - 8i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
It is just 11+8i.
The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....
it is having more advantages than 8i