Complex Numbers (1 Viewer)

[U MAD BRO]

Let Z=CIS(a) where 0 < a < pi
show that the real part of (Z-1)/(Z+1) is zero.

 

[U MAD BRO]

Bump

 

[Shadowdude]

have you tried... plugging z = cis(a) in, and then rationalising the denominator?

 

[U MAD BRO]

have you tried... plugging z = cis(a) in, and then rationalising the denominator?

Yes it works !!!
I've been substituting Z with x+iy and didn't work that way.
you get (cos^2+sin^2)-1 on the numerator which it zero
thank you

 

[Nooblet94]

You can also do it graphically. Plot the points z and 1 on the argand diagram. Then draw in z+1. You'll get a parallelogram with vertices 0, 1, z, z+1 and diagonals z+1 and z-1. Now since the diagonals of a parallelogram intersect at 90 degrees, after some rearranging you'll have arg[(z-1)/(z+1)]=90 degrees, which means it lies on the imaginary axis, so its real part is zero.

 

[U MAD BRO]

You can also do it graphically. Plot the points z and 1 on the argand diagram. Then draw in z+1. You'll get a parallelogram with vertices 0, 1, z, z+1 and diagonals z+1 and z-1. Now since the diagonals of a parallelogram intersect at 90 degrees, after some rearranging you'll have arg[(z-1)/(z+1)]=90 degrees, which means it lies on the imaginary axis, so its real part is zero.

That's what I did first.
But I couldn't show that they intersect at 90 degrees, now I just realized that you can easily prove that 0, 1, z, z+1 is a rhombus because all sides are equal (equal to 1) and one of the properties of the rhombus states that diagonals intersect at 90 degrees
Check out my new question, hopefully you can assist me with that one.

 

[Solution]

That's what I did first.
But I couldn't show that they intersect at 90 degrees, now I just realized that you can easily prove that 0, 1, z, z+1 is a rhombus because all sides are equal (equal to 1) and one of the properties of the rhombus states that diagonals intersect at 90 degrees
Check out my new question, hopefully you can assist me with that one.

Excuse my ignorance but how do the sides equal to one? I mean I know the vector from 0 to 1 is obviously 1 but what about from O to Z? Ie: lzl ?
How do you figure that out?

 

[Carrotsticks]

You can also do it graphically. Plot the points z and 1 on the argand diagram. Then draw in z+1. You'll get a parallelogram with vertices 0, 1, z, z+1 and diagonals z+1 and z-1. Now since the diagonals of a parallelogram intersect at 90 degrees, after some rearranging you'll have arg[(z-1)/(z+1)]=90 degrees, which means it lies on the imaginary axis, so its real part is zero.

Are you sure about this part?

Excuse my ignorance but how do the sides equal to one? I mean I know the vector from 0 to 1 is obviously 1 but what about from O to Z? Ie: lzl ?
How do you figure that out?

z=cis(a), so it has modulus 1.

 

[U MAD BRO]

Are you sure about this part?

You have to prove that it is a rhombus first

 

[Carrotsticks]

A rhombus IS a parallelogram, but more of a special type, so best to say 'rhombus' rather than 'parallelograms', where the diagonals aren't necessarily 90 degrees apart.

 

[seanieg89]

You can also do it graphically. Plot the points z and 1 on the argand diagram. Then draw in z+1. You'll get a parallelogram with vertices 0, 1, z, z+1 and diagonals z+1 and z-1. Now since the diagonals of a parallelogram intersect at 90 degrees, after some rearranging you'll have arg[(z-1)/(z+1)]=90 degrees, which means it lies on the imaginary axis, so its real part is zero.

You will want to be a little careful here, the arg of this fraction could also be -90, or even undefined (in the case z=1). This is the sort of diagram-dependence error that can be so easy to overlook.
As it happens, if the arg is -90 or undefined we must still lie on the imaginary axis.

 

[U MAD BRO]

You will want to be a little careful here, the arg of this fraction could also be -90, or even undefined (in the case z=1). This is the sort of diagram-dependence error that can be so easy to overlook.
As it happens, if the arg is -90 or undefined we must still lie on the imaginary axis.

Is it always true that the real part of (Z-1)/(Z+1) is equal to zero?

edit: it has to be Z = cis (0 < a < pi)