## anonymous 3 years ago express sqrt(5-3i) in the a+bi form

1. phi

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

2. anonymous

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

3. anonymous

$\sqrt{5-3i} = \sqrt{5} - \sqrt{3i}$

4. phi

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

5. anonymous

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

6. anonymous

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

7. anonymous

@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?

8. anonymous

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

9. phi

3-2i squared = 5-12i

10. anonymous

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

11. phi

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?

12. anonymous

yes i just checked i have the right problem

13. phi

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

14. anonymous

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

15. phi

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

16. phi

first choose + and find the answers then -

17. anonymous

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

18. phi