Let \(\langle\cdot,\cdot\rangle\) denote the usual inner product on \(\mathbb R^2\): \[ \langle (x_1,y_1),(x_2,y_2)\rangle=x_1x_2+y_1y_2. \] On \(\mathbb C\), define the Hermitian inner product by \[ (z,w)=z\overline w. \] Show that \[ \langle z,w\rangle =\frac12\bigl[(z,w)+(w,z)\bigr] =\operatorname{Re}(z,w), \] where \(z=x+iy\in\mathbb C\) is identified with \((x,y)\in\mathbb R^2\).
Solution
1. Write\[ z=x_1+iy_1, \qquad w=x_2+iy_2. \]Then\[ (z,w)=z\overline w =(x_1x_2+y_1y_2)+i(y_1x_2-x_1y_2). \]
Expand \[ (x_1+iy_1)(x_2-iy_2) =x_1x_2+y_1y_2+i(y_1x_2-x_1y_2). \]
2. Therefore\[ \operatorname{Re}(z,w)=x_1x_2+y_1y_2=\langle z,w\rangle. \]
The real part of the expression in 1 is \(x_1x_2+y_1y_2\), which is the Euclidean inner product under the usual identification \(\mathbb C\cong\mathbb R^2\).
3. One has\[ (w,z)=\overline{(z,w)}. \]
Indeed, \[ (w,z)=w\overline z=\overline{z\overline w}=\overline{(z,w)}. \]
4. Hence\[ \frac12\bigl[(z,w)+(w,z)\bigr]=\operatorname{Re}(z,w). \]
By 3, the left-hand side is \[ \frac12\bigl((z,w)+\overline{(z,w)}\bigr), \] which is the real part of \((z,w)\).
5. Combining the preceding identities gives\[ \langle z,w\rangle =\frac12\bigl[(z,w)+(w,z)\bigr] =\operatorname{Re}(z,w). \]
Use 2 and 4.