another complex numbers question (1 Viewer)

[ADrew]

If mod(z)=mod(w), prove that (z+w)/(z-w) is purely imaginary.
Thanks!

 

[study-freak]

Let z=rcis(alpha), w=rcis(beta) so that r=mod(z)=mod(w).
Then r cancels out from everything and multiply the numerator by the conjugate of the denominator.
Take the real part of it and show that it's 0.
=> purely imaginary

 

[ADrew]

thanks heaps! I never thought to use polar forms.

 

[Carrotsticks]

Alternatively, you can use a geometric argument.

If Z and W have the same modulus, then it means the vectors Z, W, Z+W and Z-W will draw out a rhombus (the moduli are equal, so equal length sides ie: rhombus).

However one property of the rhombus is that the diagonals are perpendicular.

Furthermore, the diagonals are the equivalent of the vectors Z+W (the long one) and Z-W (the short one).

Hence, (z+w)/(z-w) will give us some sort of 90-degree rotation (rotation can go in either direction, it doesn't matter).

But 90 degree rotation is the same as multiplying one vector by cis90 or cis(-90), both of which are purely imaginary results. We can ignore any scalars since it doesn't change the fact that they're imaginary.

Hence (z+w)/(z-w) is purely imaginary.

This is just a bit of a quick explanation to let you know that a geometric argument exists. I can explain further if needed.

 

[zeebobDD]

Alternatively, you can use a geometric argument.
I can explain further if needed.

hey carrotsticks how would you set this out in like an exam time question? with the geometric proof, like draw the diagram and show the rotation?

 

[math man]

for an exam:

given that |z|=|w|
by vector addition a rhombus will be formed (here you draw a diagram including both z+w and z-w)
therefore, the angle between diagonals is 90 degrees.
i.e. arg[(z+w)/(z-w)]= 90 degrees (using angle between two vectors)
hence, (z+w)/(z-w) is purely imaginary.

that's all you need to write which takes under 1 min