Quick question about graphing complex numbers and arguments (1 Viewer)

[ProdigyInspired]

Why does graphing a complex number i.e. z + i move it up/down accordingly to the sign of its values,

whilst an argument for arg(z+i) moves down - opposite to the sign of its imaginary values.

 

[InteGrand]

Why does graphing a complex number i.e. z + i move it up/down accordingly to the sign of its values,

whilst an argument for arg(z+i) moves down - opposite to the sign of its imaginary values.

Given a point z, the point z+i is one unit up from this in the complex plane, because the real part is unchanged but the imaginary part has increased by 1 due to adding i.

Meanwhile, for arg(z+i), this is arg(z – (-i)). Recall that to sketch arg(z - z₀) = theta (where z₀ is a fixed complex number), we shift the locus of arg(z) = theta (which is a ray starting at the origin, excluding the origin, making angle theta with the positive real axis) so that the ray starts at z₀ instead. So for arg(z+i) = theta, since it's arg(z – (-i)) = theta, we shift the ray arg(z) = theta to start at -i, which means we shift it down one unit.

 

[Ambility]

http://i.imgur.com/RlJKsj2.png

 

[uncertainty]

Its what happens when you graph an equation instead of a point?

Graph z + i when z = 1 + 3i and it is just the point moved up.

I think instead try to graph z + i = 1 + 3i and you get z = 1 + 2i which is down.

 

[InteGrand]

Its what happens when you graph an equation instead of a point?

Graph z + i when z = 1 + 3i and it is just the point moved up.

I think instead try to graph z + i = 1 + 3i and you get z = 1 + 2i which is down.

Let w equal the complex number 1 + 3i.

Then the first one is the point z = w + i, hence move up by 1 unit from w.

The second is z + i = w, which means z = w – i, so move down by one unit from w.