# 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 $z=a+bi$

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

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

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

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

Solve the following equations in $x,y$:

$z = x + iy \cdots(1)$

and $z^* = x - iy \cdots (2)$

Adding $1$ and $2$ we get $z + z^* = 2x \implies x = \frac{z + z^*}{2}$ i.e. $Re (z) = (z + z^* )/2$

Subtracting $1$ and $2$ we get $z - z^* = 2iy \implies y = \frac{z - z^*}{2i}$ i.e. $Im (z) = (z − z^*)/2i$

Let

$$z = r \exp (i \theta) = r \cos(\theta) + i r \sin(\theta)$$

With

$$z^* = r \exp (-i \theta) = r \cos(\theta) - i r \sin(\theta)$$

Then:

$$z + z^*= 2r \cos(\theta) = 2\Re (z)$$

and:

$$z - z^* = 2 i r \sin(\theta) = 2i\Im(z)$$

from which the desired result follows.

$$z=x+yi \text{ and } z^*=x-yi$$

So,

$$z+z^*=2x \to x=\frac{z+z^*}{2}\\ z-z^*=2yi \to y=\frac{z-z^*}{2i}$$