SOLUTION: Please could someone help with this question. I can't get the last part. Find real numbers a and b with a>0 such that: (a+ib)^2 = 1-1.875i So far I have got this far. a^2

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Please could someone help with this question. I can't get the last part. Find real numbers a and b with a>0 such that: (a+ib)^2 = 1-1.875i So far I have got this far. a^2      Log On


   

Question 981120: Please could someone help with this question. I can't get the last part.
Find real numbers a and b with a>0 such that:
(a+ib)^2 = 1-1.875i
So far I have got this far.
a^2-b^2 +2abi = 1-1.875i
Real: a^2-b^2=1 (1)
Imaginary: 2ab = -1.875 (2)
Divide imaginary part (2) becomes: a =-0.9375/b (3)
I got 0.9375 =15/16
Sub in (1) gives:
(15/16)^2-b^4=b^2
b^4+b^2-225/256=0
From here I can not seem to factorize. Also I tried to use the quadratic formula but that didn't seem to work either.
Thank you in advance.
The answer given in the textbook is: a=5/4 and b=-3/4




Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
%28a%2Bbi%29%5E2=1-1.875i

a%5E2%2B2abi%2Bb%5E2%2Ai%5E2

a%5E2%2Bb%5E2%2Ai%5E2%2B2abi

a%5E2%2Bb%5E2%28-1%29%2B2abi

a%5E2-b%5E2%2B2abi=1-1.875i

Corresponding positions of the terms
system%28a%5E2-b%5E2=1%2C2ab=-1.875%29

Use the second equation and substitute for b in the first equation.

a%5E2-%281.875%29%5E2%2F%284a%5E2%29=1

4a%5E4-4a%5E2-%281.875%29%5E2=0
Things from this quadratic-form equation in "a" are not very neat.
Some simplification after using formula for the general solution for quadratic equation gives highlight_green%28a%5E2=%281%2B-+sqrt%281%2B1.875%5E2%29%29%2F2%29, not yet a to first power, so not done yet..., but at least you are finding a real number for a.

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Further computing for a^2 finally showed (with handheld calculator help),
a%5E2=1.5625
and since you're interested in positive values, find
highlight%28highlight%28a=1.25%29%29.