Best Answer

Find all solutions of z4 = -8i.

Recall:

z = a + bi, the complex form

z = |z|(cos Î¸ + i sin Î¸), the polar coordinate form

So we can write:

z4 = [|z|(cos Î¸ + i sin Î¸)]4 = |z|4(cos 4Î¸ + i sin 4Î¸); -8i = 8(0 - i), and we have

|z|4(cos 4Î¸ + i sin 4Î¸) = 8(0 - i).

Thus, |z|4 = 8, so |z|= 81/4.

The angle Î¸ for z must satisfy cos 4Î¸ = 0 and sin 4Î¸ = -1.

Consequently, 4Î¸ = 3pi/2 + 2npi for an integer n, so that Î¸ = 3pi/8 + npi/2.

The different values of Î¸ obtained where 0 â‰¤ Î¸ â‰¤ 2pi are:

n = 0, Î¸ = 3pi/8 (1st quadrant)

n = 1, Î¸ = 3pi/8 + pi/2 = 7pi/8 (2nd quadrant)

n = 2, Î¸ = 3pi/8 + pi = 11pi/8 ( 3rd quadrant)

n = 3, Î¸ = 3pi/8 + 3pi/2 = 15pi/8 (4th quadrant))

Thus the solutions of z4 = -8i are

(81/4)(cos 3pi/8 + i sin 3pi8) â‰ˆ 0.6435942529 + 1.553773974 i

(81/4)(cos 7pi/8 + i sin 7pi/8) â‰ˆ -1.553773974 + 0.6435942529 i

(81/4)(cos 11pi/8 + i sin 11pi/8) â‰ˆ -0.6435942529 - 1.553773974 i

(81/4)(cos 15pi/8 + i sin 15pi/8) â‰ˆ 1.553773974 - 0.6435942529 i

Q: What is the 4th root of -8i by De morives theorem?

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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.

Related questions

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....

x2 - 2x + 65 = 0 Subtract 64 from each side: x2 - 2x + 1 = -64 (x - 1)2 = (i*8)2 where i is the imaginary square root of -1 x - 1 = +/- 8i x = 1 +/- 8i