FP1 - Complex numbers (proof) question

This is the question:

Given that

z=x+iy

and

z^2=a+ib

prove that

2x^2=\sqrt(a^2+b^2)+a

Attempt:

Expanding

z^2

and comparing coefficients:

x^2-y^2+2xyi=a+ib

x^2-y^2=a

2xy=b

I assumed there would be some substitution for y:

y = \frac{b}{2x}

y^2=\frac{b^2}{4x^2}

x^2-(\frac{b^2}{4x^2})=a

4x^4-b^2=4ax^2

which just seems like a dead end?

(edited 8 years ago)

Reply 1

zetamcfc

Original post by SkyJP

This is the question:

Given that

z=x+iy

and

z^2=a+ib

prove that

2x^2=\sqrt(a^2+b^2)+a

Attempt:

Expanding

z^2

and comparing coefficients:

x^2-y^2+2xyi=a+ib

x^2-y^2=a

2xy=b

I assumed there would be some substitution for y:

y = \frac{b}{2x}

y^2=\frac{b^2}{4x^2}

x^2-(\frac{b^2}{4x^2})=a

4x^4-b^2=4ax^2

which just seems like a dead end?

Think about Modulus. And equating real and imaginary parts.

Reply 2

TeeEm

Original post by SkyJP

This is the question:

Given that

z=x+iy

and

z^2=a+ib

prove that

2x^2=\sqrt(a^2+b^2)+a

Attempt:

Expanding

z^2

and comparing coefficients:

x^2-y^2+2xyi=a+ib

x^2-y^2=a

2xy=b

I assumed there would be some substitution for y:

y = \frac{b}{2x}

y^2=\frac{b^2}{4x^2}

x^2-(\frac{b^2}{4x^2})=a

4x^4-b^2=4ax^2

which just seems like a dead end?

no dead end
continue from there

Reply 3

Subspace

Original post by zetamcfc

Think about Modulus. And equating real and imaginary parts.

Thanks a lot, but why does this work? (don't really understand tbh)

z=x+yi

|z|=\sqrt(x^2+y^2)

|z|^2=x^2+y^2

z^2=a+bi

|z^2|=\sqrt(a^2+b^2)

x^2+y^2=\sqrt(a^2+b^2)

From before:

y^2=x^2-a

x^2+x^2-a=\sqrt(a^2+b^2

2x^2=\sqrt(a^2+b^2)+a

Reply 4

zetamcfc

Original post by SkyJP

Thanks a lot, but why does this work? (don't really understand tbh)

In what way? I would advise drawing an Argand Diagram to allow you to see what is going on.

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FP1 - Complex numbers (proof) question

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