Does $\lim_{z \to 0} \frac {Re(z^2)}{|z|^2}$ exists???

Question

does the limit exist for the following:

$$\lim_{z \to 0} \frac {Re(z^2)}{|z|^2}$$

My take: I tried to substitute $z=x+iy$ for z and then solve the limit but I get 0. According to the wolfram alpha its limit is 1; so can someone please explain this to me? What am I doing wrong here?

Thank You!!!

Answer 1

If $z \in \mathbb R$ and $z \ne 0$, then $\frac {Re(z^2)}{|z|^2}=1$

and

if $z \in i\mathbb R$ and $z \ne 0$, then $\frac {Re(z^2)}{|z|^2}=-1$

Conclusion ?

Answer 2

take $$z_n = 1/n~~~ and ~~~~w_n=1/in$$

both sequences converge to $0$ but $$ \frac {Re(z_n^2)}{|z_n|^2}=1~~~~~and~~~~~\frac {Re(w_n^2)}{|w_n|^2}=-1$$

then the convergence is impossible.

Answer 3

Assume $z \not =0:$

$z=re^{i \phi}$; $r \not = 0$.

$Re(z^2)= r^2\cos (2 \phi)$; $|z^2|=r^2.$

Consider:

$ \dfrac {Re (z^2)}{|z^2|} = \dfrac {r^2 \cos(2\phi)}{r^2}= \cos(2\phi)$;

$\phi$ is arbitrary.

Does limit $r \rightarrow 0$ exist?

Answer 4

Following your first idea, note that in algebraic form we obtain

$$ \frac {Re(z^2)}{|z|^2}=\frac{x^2-y^2}{x^2+y^2}$$

which tends to

• 0 when moving along $x=y$

• 1 when moving along $y=0$

Thus $\lim_{z \to 0} \frac {Re(z^2)}{|z|^2}$ does not exist.