√15+8i in the form a + bi, where a and b are real numbers:

Answer 1

i dont get it. is it √(15) + 8i or √(15 + 8i)

Answer 2

it's √(15 + 8i)

Answer 3

15+8i = 16+8i-1 = (4+i)^2

Answer 4

let z = 15 + 8i
argument = √(225+64) = 17
tanØ = 8/15 , Ø in quad I of the Argand plane
Ø = 28.0725°

then z = 17cis 28.0725°
then √(15 + 8i) = z^(1/2)
= √17 cis 14.03624..°
= √17(cos 14.036..° + i sin14.036..)
= 4 + i <---- primary root

other root = √17 cis(14.036.. + 180)*
= -4 - i

check:
(4+i)^2 = 16 + 8i + i^2
= 16 + 8i - 1
= 15 + 8i, my answer is correct

Answer 5

15 + 8 i = r e^iT
r = sqrt (225 + 64) =17
and
T = cos^-1 (15/ 17)
so we want the square root of 17 e^[ i cos^-1(15/17)]
= sqrt 17 * e^[ i*.5* cos^-1(15/17)]
cos^-1 (15/17) = 28 degrees
= sqrt 17 * e^[ i*.5* 28 deg]
= sqrt 17 * e^[ i*14 deg]
but e^ i A = cos A + i sin A
so
sqrt 17 [ cos 14 deg + i sin 14 deg]

Answer 6

agree with Reiny