Complex Arithmetic/Examples/3(1+i) + 2(4-3i) - (2+5i)
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Example of Complex Arithmetic
- $3 \paren {1 + i} + 2 \paren {4 - 3 i} - \paren {2 + 5 i} = 9 - 8 i$
Proof 1
\(\ds \) | \(\) | \(\ds 3 \paren {1 + i} + 2 \paren {4 - 3 i} - \paren {2 + 5 i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {3 + 3} + \paren {8 - 6 i} - \paren {2 + 5 i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren 3 + 8 - 2} + \paren {3 - 6 - 5} i\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 - 8 i\) |
$\blacksquare$
Proof 2
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 $-2 - 5 i$.
$\blacksquare$