Consider two complex numbers \({z_1}\) and \({z_2}\), given by:

\[{z_1} = {x_1} + i{y_1}, \quad {z_2} = {x_2} + i{y_2}\]

Intuitively, it seems apparent that when we add these two complex numbers, then the sum will be obtained by adding the real parts separately and the imaginary parts separately:

\[\begin{align}{z_1} + {z_2}& = \left( {{x_1} + i{y_1}} \right) + \left( {{x_2} + i{y_2}} \right)\\&= \left( {{x_1} + {x_2}} \right) + \left( {i{y_1} + i{y_2}} \right)\\& = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)\end{align}\]

Similarly, upon subtracting these two numbers, we have:

\[\begin{align}{z_1} - {z_2}&= \left( {{x_1} + i{y_1}} \right) - \left( {{x_2} + i{y_2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= \left( {{x_1} - {x_2}} \right) + \left( {i{y_1} - i{y_2}} \right)\\&= \left( {{x_1} - {x_2}} \right) + i\left( {{y_1} - {y_2}} \right)\end{align}\]

The real and imaginary parts add / subtract separately because they are in *perpendicular directions*. You will understand this better at a later stage.

Note that adding two complex numbers yields a complex number - thus, the Complex Set is closed under addition. It is also closed under subtraction.

**Example1:** Let

\[{z_1} = 2 - 3i,\,\,\,{z_2} = - 3 + 4i\]

Find \({z_1} + {z_2}\) and \({z_1} - {z_2}\).

**Solution:** We have:

\[\begin{align}{z_1} + {z_2}& = \left( {2 - 3i} \right) + \left( { - 3 + 4i} \right)\\& = \left\{ {2 + \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) + 4} \right\}\\& = - 1 + i\end{align}\]

And,

\[\begin{align}{l}{z_1} - {z_2} &= \left( {2 - 3i} \right) - \left( { - 3 + 4i} \right)\\&= \left\{ {2 - \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) - 4} \right\}\\& = 5 - 7i\end{align}\]