# Z is a complex number.?

I need some help to solve this math question.

Z is a complex number.

Z= a+bi

Z²=x+yi

Prove that 2x²=(a²+b²)^(1/2) +a

Relevance

Z²=(a+bi)(a+bi)

Z²=a²+2abi-b²

Z²=(a²-b²)+(2ab)i

x=a²-b²

2x²=2(a²-b²)²≠(a²+b²)^(1/2) + a (see, for insance, a=1 and b=1)

So either you wrote the problem wrong or this is supposed to be some kind of trick question, because the identity you're supposed to prove does not actually hold.

• Square Z:

a^2+2abi-b^2

Set it equal to Z^2:

a^2+2abi-b^2=x+iy

We know that the real parts and imaginary parts are essentially independent of each other (they satisfy equations independently) so set the real and imaginary parts equal:

a^2-b^2=x

2ab=y

So, from here, I think something in your problem is wrong. The reason I say this is because:

x^2=(a^2-b^2)^2

What you have in your problem not only has a plus sign between the two of them (which is possible but not with the other terms you have there) but also is taken the square root of (Which seems really odd). Check your problem and I'll keep an eye on the question.

• Given:

Z= a + bi

Z²= x + yi

----------------

Z= a + bi

Z²= (a + bi)(a + bi)

Z²= a² + abi + abi + bi²

Z²= a² + 2abi + bi²

But i = √(-1)

→ i² = √(-1)*√(-1)

→ i² = -1

So:

Z²= a² + 2abi - b²

x + yi= a² + 2abi - b² (match up terms).

→ x = a² - b²

→ 2x = 2(a² - b²)

→ yi = 2abi

→ y = 2ab

This does not match up with what is given in the question.

• z = a+bi

z^2 = (a+bi)(a+bi) = aa - bb + 2abi

z^2 = x+yi

thus x = aa - bb

2xx = 2(aa - bb)(aa - bb) = 2a^4 + 2b^4 -4aabb

but 2xx = (aa+bb)^(1/2)+a, therefore

2a^4 + 2b^4 -4aabb = (aa+bb)^(1/2)+a

2a^4 + 2b^4 -4aabb - a = (aa+bb)^(1/2)

 4(a^4 + b^4 - 2aabb - a)^2 = aa + bb,

ie, a and b satisfy 

• Anonymous