complex numbers (1 Viewer)

relativity1 · Nov 28, 2015

1)Suppose z is a complex number that lies on the unit circle 0≤arg(z)≤pi/2
Prove 2arg(z+1)=arg(z)

InteGrand · Nov 28, 2015

relativity1 said:
1)Suppose z is a complex number that lies on the unit circle 0≤arg(z)≤pi/2
Prove 2arg(z+1)=arg(z)

$\text{Draw the complex number z on the unit circle represented by the point P (it is in the first quadrant). Let A be the point (1,0) in the complex plane, so A represents the number 1.}$

$\text{Then} \ \overrightarrow{OQ} =\overrightarrow{OP} + \overrightarrow{OA}, \ \text{since z+1 = (z) + (1), and Q, P and A represent z+1, z and 1 respectively.$

$\text{So by the parallelogram law for addition of vectors, we have that the point Q representing the point z+1 is the point such that POAQ (in this order) is a parallelogram.}$

$\text{Now,} \ \angle POA = \arg (z) \ \text{and} \ \angle QOA = \arg (z+1). \ \text{But} \ \angle POA = 2\angle QOA \ \text{(diagonal of parallelogram POAQ bisects its angles). Hence:}$

$2\arg (z+1) = \arg(z).$

Moderator note: LaTeX input has been adjusted to make the working valid and visible.

Drongoski · Nov 28, 2015

Then z is any point that lies on the unit circle in the 1st quadrant. z+1 = z - (-1) is the vector from the point (-1,0) to any point z; arg (z+1) is the angle this vector makes with the positive x-axis and arg z is the angle the vector joining the origin to z makes with the positive x-axis. This latter is an angle on an arc at the centre of the unit circle = 2 x angle on same arc subtended at (-1,0), a point on the (same side) on the circumference.

complex numbers (1 Viewer)

relativity1

New Member

InteGrand

Well-Known Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)