Let x and y be two complex numbers. Prove the equality…
Let $x$ and $y$ be two complex numbers. Prove the equality $$ |x+y|^2 + |x-y|^2 = 2\{|x|^2+|y|^2\} $$ and give a geometric interpretation.
Explanation of Method
Write $|z|^2 = z\bar z$ and expand $(x+y)\overline{(x+y)}$ and $(x-y)\overline{(x-y)}$. The mixed terms $x\bar y$ and $y\bar x$ cancel when the two expansions are added, leaving the stated identity. Geometrically, $x$ and $y$ are two sides of a parallelogram, and $x+y$, $x-y$ are its diagonals.
Step 1 · Express modulus squared via conjugates
Recall $|z|^2=z\bar z$ for any complex $z$. So:
$$ |x+y|^2=(x+y)\overline{(x+y)}=(x+y)(\bar x+\bar y), \qquad |x-y|^2=(x-y)\overline{(x-y)}=(x-y)(\bar x-\bar y) $$
Step 2 · Expand $|x+y|^2$
$$ |x+y|^2=(x+y)(\bar x+\bar y)=x\bar x+x\bar y+y\bar x+y\bar y=|x|^2+x\bar y+y\bar x+|y|^2 $$
Step 3 · Expand $|x-y|^2$
$$ |x-y|^2=(x-y)(\bar x-\bar y)=x\bar x-x\bar y-y\bar x+y\bar y=|x|^2-x\bar y-y\bar x+|y|^2 $$
Step 4 · Add the two expansions
The cross terms $x\bar y$ and $y\bar x$ cancel:
$$ |x+y|^2+|x-y|^2 = \left(|x|^2+x\bar y+y\bar x+|y|^2\right)+\left(|x|^2-x\bar y-y\bar x+|y|^2\right) = 2|x|^2+2|y|^2 $$
$$ \boxed{|x+y|^2+|x-y|^2 = 2\{|x|^2+|y|^2\}} $$
Step 5 · Alternative proof in Cartesian coordinates
Let $x=x_1+ix_2$, $y=y_1+iy_2$ with $x_1,x_2,y_1,y_2\in\mathbb{R}$. Then:
$$ |x+y|^2=(x_1+y_1)^2+(x_2+y_2)^2, \qquad |x-y|^2=(x_1-y_1)^2+(x_2-y_2)^2 $$
Adding, the cross terms $2x_1y_1$, $2x_2y_2$ cancel between the two:
$$ |x+y|^2+|x-y|^2 = 2x_1^2+2x_2^2+2y_1^2+2y_2^2 = 2(x_1^2+x_2^2)+2(y_1^2+y_2^2) = 2|x|^2+2|y|^2 $$
which confirms the result.
Step 6 · Geometric picture
View $x$ and $y$ as position vectors $\vec{OA}=x$ and $\vec{OB}=y$ from the origin $O$. Complete the parallelogram $OACB$ with $C$ at position $x+y$. Then:
- Sides: $\vec{OA}=x$, $\vec{OB}=y$, $\vec{BC}=x$, $\vec{AC}=y$ — lengths $|x|,|y|,|x|,|y|$.
- Diagonal $OC$: $\vec{OC}=x+y$, length $|x+y|$.
- Diagonal $AB$: $\vec{AB}=y-x$, length $|y-x|=|x-y|$.
Parallelogram $OACB$: sides $|x|$, $|y|$; diagonals $|x+y|$ (dashed red) and $|x-y|$ (dashed purple).
Step 7 · Statement of the parallelogram law
$$ \boxed{\text{diagonal}_1^2+\text{diagonal}_2^2 = 2\left(\text{side}_1^2+\text{side}_2^2\right)} $$
In any parallelogram, the sum of the squares of the diagonals equals the sum of the squares of all four sides (since opposite sides are equal, $|x|^2+|y|^2+|x|^2+|y|^2=2(|x|^2+|y|^2)$).
Step 8 · Special cases
| Case | Check |
|---|---|
| $x\perp y$ (rectangle) | $|x+y|^2=|x-y|^2=|x|^2+|y|^2$, sum $=2(|x|^2+|y|^2)$ ✓ |
| $y=x$ | $|2x|^2+0=4|x|^2=2(|x|^2+|x|^2)$ ✓ |
| $y=-x$ | $0+|2x|^2=4|x|^2=2(|x|^2+|x|^2)$ ✓ |
Step 9 · Numerical verification
Let $x=3+4i$, $y=1+2i$.
$$ x+y=4+6i \Rightarrow |x+y|^2=16+36=52, \qquad x-y=2+2i \Rightarrow |x-y|^2=4+4=8 $$
Left side: $52+8=60$. Right side: $|x|^2=25$, $|y|^2=5$, so $2(25+5)=60$. Both sides equal $60$. ✓
Result
Geometric interpretation (Parallelogram Law): in any parallelogram, the sum of the squares of the lengths of the two diagonals equals the sum of the squares of the lengths of all four sides. This identity also characterizes norms that arise from an inner product — here $\langle x,y\rangle=\text{Re}(x\bar y)$ on $\mathbb{C}\cong\mathbb{R}^2$.