Prove $Re (z) = \frac{z + z^* }{2}$ and $Im (z) = \frac{z − z^*}{2i}$ In texts on complex numbers I often see an exercise that asks to prove the following:

$$Re (z) = \frac{z + z^*}{2}$$
$$Im (z) = \frac{z − z^*}{2i}$$
where $z = x + iy$ and $z^* = x - iy$

I understand the meaning of complex numbers, but can't seem to find a path to proving these two identities.
Any insight would be appreciated.

Let $z=a+bi$

$Re(z)=\frac{z+z^*}{2}$

$$\left(\frac{1}{2} \right)(z+z^*)=\left(\frac{1}{2} \right)((a+bi)+(\overline{a+bi}))$$
$$=\left(\frac{1}{2} \right)((a+bi)+({a-bi}))=
\left(\frac{1}{2} \right)(2a)=a=Re(z)$$

$Im(z)=\frac{(z-z^*)}{2i}$

$$\left(\frac{1}{2i} \right)(z-z^*)=\left(\frac{1}{2i} \right)((a+bi)-(\overline{a+bi}))$$
$$=\left(\frac{1}{2i} \right)((a+bi)-({a-bi}))=\left(\frac{1}{2i} \right)(2bi)=b=Im(z)$$
