Solving Inequality Problem: 0<|z|<1 and z_1, z_2

[Garret122]

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

 

[berkeman]

Sorry, what is a?

 

[SGT]

I suppose that the first equation should be 0 <|a|<1.
The second term of the equations for $$z_1$$ and $$z_2$$ has the same sign as $$\frac{1}{a}$$.
When both terms have the minus sign, you are adding the moduli and clearly the modulus of the result is greater than 1.When the first is negative and the second positive you subtract the moduli. It remains to show that the modulus of the result is smaller than 1.