Prove the identity (complex numbers)

Question

The identity:

$(|Z|^2 - 1)^2 + (2ReZ)^2 = |Z^2+1|^2$

Say $Z = x + yi$

For the left side I simplified to $((\sqrt{(x^2+y^2)})^2 - 1)^2 = ... = x^4 + y^4 + 2x^2y^2 +2x^2 -2y^2 +1$

For the right side I have:

(since $Z^2 = x^2 + y^2$, therefore:)

$|Z^2+1|^2 = |x^2 + y^2 + 1|^2 =...= x^4 + y^4 + 2x^2y^2 + 2x^2 + 2y^2 +1$

Did I do everything correctly? I got $-2y^2$ for the left side, but $+2y^2$ for the right side. That means identity is incorrect?

Answer 1

You're on the right track

• $(|z|^2-1)^2+(2\mathfrak{R}(z))^2=(x^2+y^2-1)^2+(2x)^2$
• $|z^2+1|^2=|x^2-y^2+1+2xyi|^2=(x^2-y^2+1)^2+(2xy)^2=(x^2-(y^2-1))^2+(2xy)^2$

We can now compare: $$(x^2+(y^2-1))^2+(2x)^2=(x^2)^2+2(x^2)(y^2-1)+(y^2-1)^2+(2x)^2\\(x^2-(y^2-1))^2+(2xy)^2=(x^2)^2-2(x^2)(y^2-1)+(y^2-1)^2+(2xy)^2$$

Answer 2

Yes, you are doing it right. Note that:

• $\bigl\lvert(x+yi)^2+1\bigr\rvert^2=\bigl((x^2-y^2)+1\bigr)^2+(2xy)^2=x^4+y^4+2x^2-2y^2+2x^2y^2+1$;
• $\bigl(\lvert x+yi\rvert^2-1\bigr)^2=(x^2+y^2-1)^2=x^4+y^4-2x^2-2y^2+2x^2y^2+1$.

If you add $4x^2\bigl(=(2x)^2\bigr)$ to the second expression, you get the first one. So…

Answer 3

Recall that $$ |Z|^2=Z\bar{Z}\qquad\text{and}\qquad\operatorname{Re}Z=\frac{Z+\bar{Z}}{2} $$ Therefore \begin{gather} (|Z|^2-1)^2=(Z\bar{Z}-1)^2=Z^2\bar{Z}^2-2Z\bar{Z}+1\\[6px] (2\operatorname{Re}Z)^2=(Z+\bar{Z})^2=Z^2+2Z\bar{Z}+\bar{Z}^2 \end{gather} Hence the left hand side becomes $$ Z^2\bar{Z}^2-2Z\bar{Z}+1+Z^2+2Z\bar{Z}+\bar{Z}^2= Z^2\bar{Z}^2+Z^2+\bar{Z}^2+1=(Z^2+1)(\bar{Z}^2+1)=|Z^2+1| $$

Answer 4

A simpler way:

Remember that $|z|^2=z\bar z$ and $z+\bar z=2\operatorname{Re}z$. Now, the l.h.s. transforms as \begin{align} (|Z|^2 - 1)^2 + (2\operatorname{Re}Z)^2&=(z\bar z-1)^2+(z+\bar z)^2\\ &=(z^2\bar z^2-2z\bar z+1)+(z^2+2z\bar z+\bar z^2)\\ &=z^2\bar z^2+z^2+\bar z^2+1\\ &=(z^2+1)(\bar z^2+1)\quad\text{by a high-school identity.}\\ &= (z^2+1)(\overline{\smash{z^2}+1\mathstrut}) \end{align}

Answer 5

A trick to realize is that $(a - b)^2 + 4ab = (a+b)^2$

Just continue

$(|Z|^2 - 1)^2 + (2ReZ)^2 =$

$((x^2 + y^2) - 1)^2 + 4x^2 = $

$(x^2 + y^2)^2 - 2(x^2 + y^2) + 1 + 4x^2 =$

$(x^2 + y^2)^2 + 2(x^2 - y^2) + 1 $.

And

$ |Z^2+1|^2= |(x^2 - y^2) + 2xy i + 1|^2 =$

$ (x^2 - y^2 + 1)^2 + 4x^2y^2 = $

$(x^2 - y^2)^2 + 2(x^2 - y^2) + 1 +4x^2y^2 = $

$(x^2 + y^2)^2 + 2(x^2 - y^2) + 1$