Let's consider the following complex number: $z= 3+2i$, then you calculate $r (\arg z)$ by: $r=\sqrt {3^2+2^2}=\sqrt {13}$
But why don't you use this calculation: r=$\sqrt {3^2+(2i)^2}=\sqrt {3^2+4i^2}=\sqrt {9-4}=\sqrt{5}$?
Isn't the "height" $2i$, which is an imaginary number? Why can you simply convert it to a real number "$2$"?
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Nobody is “converting” $2i$ to $2$. The idea of $r$ is that it is the distance from $z$ to the origin. And if $z=a+bi$ (with $a,b\in\mathbb R$), this distance is $\sqrt{a^2+b^2}$.
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The $r$ is to indicate the radial length of the vector that the arguement is to indicate the direction it is pointing at. Th length of the vector is attained by using the Pythagreon theorem.
If we had a vector $1+2i$ and found $r$ as you want to, $$r=\sqrt{1-4}=\pm i\sqrt{3}$$ this does not indicate the length of the vector since $\sqrt{3}<2$
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when we represent a complex number in a+ib both a and b are real. the Imaginary part of z is usually represented on the y-axis of the complex plane.i.e. the real number b. Like @Andrew Li rightly mentioned that iota just tells us that we are dealing with the imaginary/complex plane. In fact all the equations of different loci can be changed into equations which are analogous to similar curves on the x-y plane by using the z=x+iy form.
No, its height is $2$, the real part of $2i$. $i$ has no height. We are moving $2$ units on the imaginary axis with $2i$.
The $i$ tells us we're on the imaginary plane, and that's just how we plot it on the imaginary plane. Applying the Pythagorean theorem to complex numbers, their magnitude, or distance from the origin, is the square root of the sum of the squares of the real parts.