Linear Combination of Non-Parallel Complex Numbers is Zero if Factors are Both Zero
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Theorem
Let $z_1$ and $z_2$ be complex numbers expressed as vectors such taht $z_1$ is not parallel to $z_2$.
Let $a, b \in \R$ be real numbers such that:
- $a z_1 + b z_2 = 0$
Then $a = 0$ and $b = 0$.
Proof
Suppose it is not the case that $a = b = 0$.
Then:
\(\ds a z_1 + b z_2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a \paren {x_1 + i y_1} + b \paren {x_2 + i y_2}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a x_1 + b x_2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds a y_1 + b y_2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a x_1\) | \(=\) | \(\ds -b x_2\) | |||||||||||
\(\ds a y_1\) | \(=\) | \(\ds -b y_2\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \dfrac {y_1} {x_1}\) | \(=\) | \(\ds \dfrac {y_2} {x_2}\) |
and $z_1$ and $z_2$ are parallel.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $9$