Question

express sqrt(5-3i) in the a+bi form

Answer 1

you could try using
\[ (a+bi)^2 = 5-3i \]
this means
\[ a^2-b^2 + 2abi = 5 -3i\]
equate real with real, and imag with imag. solve for a and b

Answer 2

by equate do you mean say a2^b2 = 5 and 2abi = -3i? as in set up two seperate equations?

Answer 3

\[\sqrt{5-3i} = \sqrt{5} - \sqrt{3i}\]

Answer 4

a^2 - b^2 = 5 and 2abi = -3i
yes

Answer 5

@FaisalAyaz i don't think it does. \[\sqrt{a + b} \neq \sqrt{a}+\sqrt{b}\]

Answer 6

Yes.It's Wrong.I Forgot....sorry

Answer 7

@phi for the a^2 - b^2 = 5 part i got a = +/- 3, and b = +/-2 (by guess and check, i'm not sure how to do it algebraically)
for the 2abi = -3i i got ab = -3/2
i'm slightly stuck. do i try and solve a system of equations now since i have two equations, or somehow plug the values back in, or what?

Answer 8

@phi i got 3-2i and -3+2i as my answers. is that right?

Answer 9

3-2i squared = 5-12i

Answer 10

then is it wrong that a and b are (3, -2) or (-3, 2)?

Answer 11

I am getting some very ugly numbers for this problem.
see
http://www.wolframalpha.com/input/?i=%28sqrt%28+0.5%285%2Bsqrt%2834%29%29%29-3%2F%282*sqrt%28+0.5%285%2Bsqrt%2834%29%29%29%29*i%29%5E2

Are you sure you have the correct problem?

Answer 12

yes i just checked i have the right problem

Answer 13

you can get the ugly result by saying
2ab= -3
b= -3/(2a)
b^2 = 9/(4a^2)

sub into
a^2 - b^2 =5
a^2 - 9/(4a^2) = 5
multiply by 4a^2
4a^4 -20a^2 -9 =0

or
a^4 -5a^2 -9/4 - 0

let x= a^2, so you get a quadratic
x^2 - 5x -9/4=0

use the quadratic formula to find x. then take the square root of x to find a
then solve for b

Answer 14

what bout the plus or minus part? i have [5 +/- sqrt(34)]/[2] - b^2 = 5

Answer 15

the ± give you different answers.
(square roots will have 2 answers)

Answer 16

first choose + and find the answers
then -

Answer 17

okay. i have to go now. i might sign on later, after i work on the problem some more. thank you. bye.

Answer 18

remember you can use wolfram to check your answers