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Consider the complex numbers in the following picture, as well as the line segments connecting them to the origin:

Here's a list of pairwise sums of the conjugates of these complex numbers:

\overline{z}_1+\overline{z}_2, \overline{z}_1 + \overline{z}_3, \overline{z}_1 + \overline{z}_4, \overline{z}_2 + \overline{z}_3, \overline{z}_2 + \overline{z}_4, \overline{z}_3 + \overline{z}_4.\

Find the number of the quadrant each of these pairwise sums is in, and answer with the ordered list, such that your first number corresponds to the quadrant that \overline{z}_1+\overline{z}_2 is in, your second number corresponds to the quadrant that \overline{z}_1+\overline{z}_3 is in, etc.

RektTheNoob Dec 16, 2018

#1**+2 **

"*Here's a list of pairwise sums of the conjugates of these complex numbers:*

*\(\overline{z}_1+\overline{z}_2, \overline{z}_1 + \overline{z}_3, \overline{z}_1 + \overline{z}_4, \overline{z}_2 + \overline{z}_3, \overline{z}_2 + \overline{z}_4, \overline{z}_3 + \overline{z}_4\)*

*Find the number of the quadrant each of these pairwise sums is in, and answer with the ordered list, such that your first number corresponds to the quadrant that \(\overline{z}_1+\overline{z}_2\) is in, your second number corresponds to the quadrant that \(\overline{z}_1+\overline{z}_3\)*

*is in, etc.*"

Add two of the given complex numbers, \(z_1+z_2\) say, and reflect the result in the x-axis (i.e. the Real axis).

So, for example, it looks like \(z_1+z_2\) when added will result in a complex number in the 2nd quadrant. Reflect this in the x-axis and the result will be in quadrant 3. i.e. \(\overline{z}_1+\overline{z}_2\)will be in quadrant 3.

Repeat this process for all the specified pairs.

Alan Dec 16, 2018

#3**+2 **

"*why do we reflect across the x axis????*"

Let \(z=p+iq\) , then \(\overline{z}=p-iq\)

So, if z is the resultant obtained by adding two complex numbers, the real part of the complex conjugate of z has the same magnitude and the same sign, but the imaginary part of its complex conjugate has the same magnitude but opposite sign i.e. its complex conjugate is just a reflection in the real (or x) axis.

Of course the problem could be tackled by finding both com plex conjugates of a pair and then simply adding them, but the result would be the same.

Alan
Dec 16, 2018