conjugate property (1 Viewer)

[jnney]

if |z|=|w|, prove that (z+w)/(z-w) is purely imaginary. be drawing a suitable diagram, give a geometrical interpretation if the result.

 

[math man]

this question is quickly solved geometrically by using vector addition.....as |z| = |w| we will form a rhombus and hence the diagonals z-w and z+w intersect at 90 degrees....therefore
arg[(z+w)/(z-w)] is 90 and hence (z+w)/(z-w) is purely imaginary....to prove this algebraicly....times top and bottom by complex conjugate of denominator, but then relise that the number of the
bottom is purely real using z x conjugate z = |z|^2 and so you dont expand the bottom...for the top you expand it out and use the fact |z|=|w|..after you have simplyfied this as much as you can
you sub in z= x+iy and w=a+ib on the top only and then it will be purely imaginary...sorry i am just tired so i ceebs typing it all up

 

[Carrotsticks]

as |z| = |w| we will form a parallelogram

I think you mean a rhombus. The diagonals of a parallelogram do not intersect at 90 degrees.

Good geometric proof though.

 

[Carrotsticks]

Oh and I forgot to mention that Arg(z+w)/(z-w) can actually have 2 values, depending on how you draw it. If Arg(z-w) < Arg(z+w) , then indeed you will get 90 degrees. However if Arg(z+w) < Arg (z-w), then you will in fact get -90 degrees, both of which are still purely imaginary.

Just pointing out this little detail just in case your school is really picky.

 

[math man]

I think you mean a rhombus. The diagonals of a parallelogram do not intersect at 90 degrees.

Good geometric proof though.

yeh i did...i guess doing this post at 3 am was not a good idea