For any two complex number `z_1a n d\ z_2` prove that: `|z_1-z_2|lt=|z_1|+|z_2|`

Explanation

A complex number can be represented as a vector with the real part being the magnitude of the component along the `X` axis and the imaginary part being the magnitude of the component along the `Y` axis.


Let `Z_1` and `Z_2` be represented by the vectors `\vec P` and `\vec Q` respectively.


Then, we need to prove that,
`|\vec P-\vec Q| \leq|\vec P|+|\vec Q|`
From the parallelogram law of vector addition, we get,
`|\vec P-\vec Q|^2=|\vec P|^2+|\vec Q|^2-2|\vec P||\vec Q| \cos \theta`
where `\theta` is the angle between the two vectors.
`\Rightarrow \quad|\vec P-\vec Q|^ 2 \leq|\vecP|^2+|\vec Q|^2+2|\vec P||\vec Q|`
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