1
$\begingroup$

I manipulated the series to $\sum_{n=0}^{\infty} \left( \frac{1}{1-(-1/z)} \right)^n$, which converges for $|-1/z|<1$ by geometric series. Then solving for $z$, I obtained $z>(1/\bar{z})$.

Is this correct? I was expecting some numerical values, so I'm not sure about my answer.

$\endgroup$
3
$\begingroup$

Let me suggest another approach. Put $w=\dfrac{z}{z+1}$. Where will $$\sum_{n=0}^\infty w^n$$ converge? What does this tell you about the possible domain of $z$?

ADD

I am getting $$2\Re z+1>0$$ Note you cannot really use complex numbers in an inequality, since they are not ordered in any way. However, from $|z|<|1+z|$ you get the equivalent $$z\bar z<(1+z)(1+\bar z)$$ since everything is real. This gives $$z\bar z <1+(z+\bar z)+z\bar z$$ which is the same as $$0<1+2\Re (z)$$

$\endgroup$
  • $\begingroup$ $z > 1-\bar{z}$? $\endgroup$ – AlanH May 26 '13 at 22:36
  • $\begingroup$ @AlanH I added something. $\endgroup$ – Pedro Tamaroff May 26 '13 at 22:40
  • $\begingroup$ yeah, I got the same thing, but I just didn't express the final step like that. Does it matter though? $\endgroup$ – AlanH May 26 '13 at 22:46
  • 1
    $\begingroup$ @AlanH What matters is that complex numbers aren't ordered, so $<$ makes no sense for complex numbers, so you ought to express $|z|<|z+1|$ in terms of the real and imaginary parts of $z$. $\endgroup$ – Pedro Tamaroff May 26 '13 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.