Solving Inequality Problem: 0 < |z| < 1

[Garret122]

Sorry that i posted in the wrong topic, I'm kind of new here :D
Hi this is my problem:

if 0<|z|<1 and z_1 = -1/a - ((1-a^2)^(1/2))/a
z_2 = -1/a + ((1-a^2)^(1/2))/a
Then it is clear to me that |z_1|>1 since using triangle inequality we get that |z_1| =| -1/a - ((1-a^2)^(1/2))/a | >= |1/a| + something smaller than one but positiv, and since |1/a| >1 then |z_1| > 1

But how to prove |z_2| < 1 since bye triangle inequality we kind of get the same thing |z_2| = | -1/a + ((1-a^2)^(1/2))/a | >= |1/a|+ |((1-a^2)^(1/2))/a| > 1 ? This doesn't make sense at all!

Please help me, i need this to a problem on an integral in complex analysis, which I'm preparing for my exam ;)

thank you for your time!
Garret

 

[HallsofIvy]

What exactly did you learn as "the triangle inequality"? Most people learn it as $|a+ b|\le |a|+ |b|$. From that, if we let a= x- y, b= y we get
$|x-y+y|= |x|\le |x-y|+ |y|$ so that $|x-y|\ge |x|- |y|$. That second inequality is what you used. To prove the second part use the inequality $|a+b|\le |a|+ |b|$.