- #1

Garret122

- 3

- 0

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