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:

Complex-Addition-(3(1+i))+(2(4-3i))-(2+5i)--1.png


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

Complex-Addition-(3(1+i))+(2(4-3i))-(2+5i)--2.png

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 $-2 - 5 i$.

$\blacksquare$