Question

if a and b are real numbers such that (a+ib)^2 =4, verify that a^4 + a^2-16=0

Answer 1

You don't need to worry about b, because the only square roots of 4 are purely real: 2 and -2. So, a = 2 or -2, and b=0. I'm assuming you actually meant \[a^4+b^2-16=0\], because the equation you wrote isn't actually true.

Answer 2

thats the thing, it's a past exam paper. i copied the question as is

Answer 3

Looks like a mistake in the exam, then.

Answer 4

and all you have said is true, b =0

Answer 5

but then, one more question, if that is \[b ^{2}\] = 0,
then the equation in the exam is correct because \[2^{4} + 2^{2} - 20 = 0. \]

Answer 6

I thought it was \[a^4+b^2-16=0\]

Answer 7

so how do you verify that its correct, i can tell how you got b= 0, and a=+/- 2, but how do you get to that equation? the equation in the exam is \[a ^{4}+ a ^{2} -20 =0\]

Answer 8

@srossd

Answer 9

Oh, if it is 20, then there's no mistake. Just plug in a = 2 and a = -2, and verify that the equation holds true for both.

Answer 10

and i just did that haha ha. thanx!