21Let T= \frac{z_1}{z_2},z_2 \ne 0 ,|z_1+z_2|>|z_1-z_2|\iff|T+1|>|T-1|\iff|T+1|^2>|T-1|^2 \iff (T+1)\overline{(T+1)}> (T-1)\overline{(T-1)}\iff 2(T+ \overline{T}) >0 \iff \Text{Re} (T)>0 \iff -\frac \pi 2 < \text{arg}(T)< \frac \pi 2
if |z1 + z2| > |z1 - z2| then prove that -pi/2 < arg(z1/z2) < pi/2
-
UP 0 DOWN 0 0 3
3 Answers
21
341Let z_0 = \frac{z_1}{z_2}
Then the region \left|z_0 +1 \right|>\left|z_0 -1 \right| consists of those numbers whose distance from -1 is greater than that from 1. This are the numbers that lie to the right of y-axis i.e. 1st and 4th quadrants and the conclusion follows
36Thanks, this is what i wanted...
but geometry kills it more easily....!!
case 1> If the angle between z1 and z2 is 90° then |z1 + z2| = |z1 - z2| so pi/2 is not possible
case 2> When its greater than pi/2
then |z1 - z2| > |z1 + z2| so a contradiction again...
case 3> When its less than -pi/2 ... the case is similar to case 2