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If I simply put the points $(i,0)$ and $(0,1)$ into the equation $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ I get $$Distance=\sqrt{(0-i)^2+(1-0)^2}$$ $${Distance=\sqrt{(-i)^2+(1)^2}}$$ $$Distance_=\sqrt{-1+1}$$ $$Distance=\sqrt{0}$$ $$Distance=0$$ So does this mean that the distance between the points $(i,0)$ and $(1,0)$ is $0$ or does the formula $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ not work for imaginary numbers?

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  • $\begingroup$ No, you should define the metric a different way. $\endgroup$ – Botond Dec 3 '17 at 9:43
  • $\begingroup$ This is one of the possible representation of a Minkowski space. en.wikipedia.org/wiki/Minkowski_space $\endgroup$ – Emilio Novati Dec 3 '17 at 15:02
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Jan 17 '18 at 22:56
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$$ distance = \sqrt{|x_1 - x_2|^2 + |y_1 - y_2|^2 } $$

PS. Usually, we define norm in linear space using scalar product as $\| x \| = \sqrt{(x, \, x)}$. And standard scalar product in complex spaces is defined as $$ (x, \, y) = \sum x_i \bar{y}_i .$$

Notice the bar over $y_i$.

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The usual distance formula doesn't work for complex numbers. For even better example take $(i,0)$ and $(0,0)$. The distance formula would then yield $i$. Can a distance be a complex number?

You can modify the distance formula a bit, writing

$$\text{Distance} = \sqrt{|x_1 - x_2|^2 + |y_1 - y_2|^2}$$

This will work, where $|x_1 - x_2|$ is the distance in the complex plane. In other words if $x_1 = a + ib$ and $x_2 = x+iy$ then we have $|x_1 - x_2| = |(a-x) + i(b-y)| = \sqrt{(a-x)^2 + (b-y)^2}$

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$(i,0)$ is not a point in $\mathbb R^2$. If you're looking at the representation of complex numbers on the Cartesian plane, ie. $z=(\mathrm{Re}(z), \mathrm{Im}(z)) \in \mathbb R^2$, then you want the point $(0,1)$, from which,

$$\mathrm{Distance} = \sqrt{(0-1)^2 + (1-0)^2} = \sqrt 2$$

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  • $\begingroup$ This is the correct answer IMHO. I believe the OP is mixing up the Argand plane and the coordinate plane together. $\endgroup$ – Gaurang Tandon Dec 3 '17 at 9:58
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$(a,b)$ refers to points on the $x$–$y$ plane, not on the real–imaginary plane. Thus, a point $(i,0)$ doesn’t exist or make sense.

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  • $\begingroup$ It does make sense in two-dimensional linear space over field $\mathbb{C}$. $\endgroup$ – Pavel Ievlev Dec 3 '17 at 12:08
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in the complex plane "i" correspond to the point (0,1)

this you can apply the formula for the two points (1,0) and (0,1)

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