I asked time ago about the limit of a complex number $z$ over its conjugate, as $z$ goes to infinity.

Now I have a strategy, it is to convert $z$ to trigonometric form, and the limit depends uniquely on the norm of $z$, which can be eliminated from the fraction, so you got a fraction $\cos\theta+i\sin\theta$ over $\cos\theta-i\sin\theta$ as the norm of $z$ goes to infinity.

Since the function doesn't depend on that, does the limit exist?

Thank you!

  • $\begingroup$ I just noted that in the previous question from the OP there were already enough information to answer the question, so probably this one should be closed. $\endgroup$ – enzotib Aug 14 '12 at 8:22
  • $\begingroup$ In addition to the answer to the previous question, I wrote a comment that in addition more or less writes down a formal $\epsilon-\delta$ proof that the limit does not exist. $\endgroup$ – André Nicolas Aug 14 '12 at 9:02
  • $\begingroup$ Up to normalization this looks like squaring: $\frac{z}{\bar{z}} = \frac{re^{i \theta}}{re^{- i \theta}} = e^{i 2\theta}$... $\endgroup$ – uncookedfalcon Aug 14 '12 at 11:22

Let's be explicit: consider $z=x+iy$ . Then we have $$ \frac{z}{\bar z} = \frac{x+iy}{x-iy} = \frac{x^2-y^2+2ixy}{x^2+y^2} = 1 + \frac{-2y^2+2ixy}{x^2+y^2} = -1 + \frac{2x^2+2ixy}{x^2+y^2} ~~, $$ the latter two appearing by adding and subtracting in one case $x^2$, in the other $y^2$ from the top of the fraction.

When we say $z\to\infty$, as you mentioned this means that the norm of $z$ must grow without bound, and so $x^2+y^2 \to \infty$. However, there are many, many ways for this quantity to grow to infinity; for instance, take the simple cases of traveling to the right along the $x$-axis (so $y=0$, $x\to\infty$) and traveling up the $y$-axis (so $x=0$, $y \to\infty$) . Clearly both satisfy $x^2+y^2 \to \infty$ .

But notice that $$ \lim_{y=0,x\to\infty} 1 + \frac{-2y^2+2ixy}{x^2+y^2} = 1 ~~, $$ while $$ \lim_{x=0,y\to\infty} -1 + \frac{2x^2+2ixy}{x^2+y^2} = -1 ~~. $$ Hence, the limit cannot exist. The axes are just the simplest choices -- taking different paths to infinity can give you different limits.


The limit depends on $\theta$, or in other word it is different for each direction you reach infinity. Conclusion: the limit does not exist.


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.