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Properties of Modulus

Let x1 and ...

Question

Let $x_{1}$ and $x_{2}$ be two complex numbers such that $| x_{1} + x_{2} | = | x_{1} | + | x_{2} |$ . show that arg $(x_{1}) - arg (x_{2}) = 0$

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Solution

Let $x_{1} = a_{1} + i b_{1}$
$x_{2} = a_{2} + i b_{2}$
$| x_{1} | + | x_{2} | = | x_{1} + x_{2} |$
$\sqrt{a_{1}^{2} + b_{1}^{2}} + \sqrt{a_{2}^{2} + b_{2}^{2}} = \sqrt{(a_{1} + a_{2})^{2} + (b_{1} + b_{2})^{2}}$
square both sides
$2 \sqrt{(a_{1}^{2} + b_{1}^{2}) (a_{2}^{2} + b_{2}^{2})} = 2 a_{1} a_{2} + 2 b_{1} b_{2}$
$\frac{b_{1}}{a_{1}} = \frac{b_{2}}{a_{2}}$
$a r g (x_{1}) = a r g (x_{2}) = \frac{b_{1}}{a_{1}} = \frac{b_{2}}{a_{2}}$
$a r g (x_{1}) - a r g (x_{2}) = 0$

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