# Complex function limit

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!

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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. – enzotib Aug 14 '12 at 8:22
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. – André Nicolas Aug 14 '12 at 9:02
Up to normalization this looks like squaring: $\frac{z}{\bar{z}} = \frac{re^{i \theta}}{re^{- i \theta}} = e^{i 2\theta}$... – 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.