Complex Arithmetic/Examples/3(1+i) + 2(4-3i) - (2+5i)/Proof 2

Example of Complex Arithmetic

$3 \paren {1 + i} + 2 \paren {4 - 3 i} - \paren {2 + 5 i} = 9 - 8 i$

We have:

$\ds 3 \paren {1 + i}$

$=$

$\ds 3 + 3 i$

$\ds 2 \paren {4 - 3 i}$

$=$

$\ds 8 - 6 i$

$\ds -\paren {2 + 5 i}$

$=$

$\ds -2 - 5 i$

These can be depicted in the complex plane as follows:

To find the required sum, proceed as in the following diagram:

Construct $8 - 6 i$ with its initial point placed at the terminal point of $3 + 3 i$.

Construct $-2 - 5 i$ with its initial point placed at the terminal point of this instance of $8 - 6 i$.

The required resultant $3 \paren {1 + i} + 2 \paren {4 - 3 i} - \paren {2 + 5 i}$ is therefore represented by the terminal point of $-2 - 5 i$.

$\blacksquare$

1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $61 \ \text {(d)}$