ianc said:
Hi I'm just not sure how to approach this question:
If |z|=|w|, prove that (z+w)/(z-w) is purely imaginary. Draw a diagram to give a geometrical interpretation of the result.
Thanks!
hey hey ianc
welcome to the BOS forums
okay, for this one, u must consider the solution diagrammatically.
consider any z and w on the argand diagram. we know that |z| = |w|, so their modulus are equal. they are of equal lengths.
NOW, draw z+w and z-w. notice that the vertices at O, z, w and z+w form a rhombus, since all the sides of quadrilateral formed are equal. it is important to realise that the diagonals are
perpendicular. HAHA!
realise that the diagonals are represented by the complex numbers
z+w and
z-w HAHA!
this might help --> since its a rhombus, we know that |z+w| = |z-w|
we also know that arg(z+w) - arg(z-w) = 90 degrees
THESE two bits of information AND the ability to express complex numbers by the mod-arg form, it develops into this equation:
(z+w) = cis(90
o)(z-w)
therefore, (z+w)/(z-w) = i --> which is imaginary
of course, u wouldnt set our ur working out this way, im only pointing out the important parts u need to get in order to answer the question.
