(a+bi)^2=4+4sqrt3i

I got

a^2-b^2+2abi=4+4sqrt3i

a^2=b^2+4

a=sqrt(b^2+4)

2ab=4sqrt3

a=4sqrt3/2b

I can't seem to get the answer when I compare the values

The answer given for a is +/- sqrt 6 and b is +/- sqrt2

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- Aug 11th 2011, 05:25 AMfuzzyFind a and b for complex number
(a+bi)^2=4+4sqrt3i

I got

a^2-b^2+2abi=4+4sqrt3i

a^2=b^2+4

a=sqrt(b^2+4)

2ab=4sqrt3

a=4sqrt3/2b

I can't seem to get the answer when I compare the values

The answer given for a is +/- sqrt 6 and b is +/- sqrt2 - Aug 11th 2011, 05:36 AMHallsofIvyRe: Find a and b for complex number

I would avoid square roots and start from this. Now so that becomes . Multiply both sides by to get of .

That is a quadratic equation of - and easily factorable. You will find two values for so 4 different values for b.

Quote:

I can't seem to get the answer when I compare the values

The answer given for a is +/- sqrt 6 and b is +/- sqrt2

- Aug 11th 2011, 05:43 AMProve ItRe: Find a and b for complex number
This might be easiest to solve using polars.

, so