Limit of complex function

**mukmar** · November 14th 2010, 04:04 PM

Find $\displaystyle\lim_{z \to 0} \frac{\bar{z}^2}{z}$

Attempt: I thought that this limit might not exist since if you let $z = x + iy$ , then when approaching from y-axis, the limit equals $iy$ and when approaching from the x-axis, the limit equals $x$ .

However the solution says that limit is actually 0. I'm not sure how to get this result.

Any suggestions?

**FernandoRevilla** · November 14th 2010, 04:37 PM

Use $|\bar{z}^2/z|=|z|$ .

Regards.

**mukmar** · November 14th 2010, 06:30 PM

Is there a limit law that says that the limit of the modulus is related to the limit of the function in any way?

**chisigma** · November 14th 2010, 07:09 PM

Originally Posted by mukmar

Find $\displaystyle\lim_{z \to 0} \frac{\bar{z}^2}{z}$

Attempt: I thought that this limit might not exist since if you let $z = x + iy$ , then when approaching from y-axis, the limit equals $iy$ and when approaching from the x-axis, the limit equals $x$ .

However the solution says that limit is actually 0. I'm not sure how to get this result.

Any suggestions?

Setting $z=x + i\ y$ You obtain...

$\displaystyle \frac{\bar{z}^{2}}{z} = \frac{x^{2} - y^{2} -2 i x y}{x+i\ y}= \frac{x^{3} -3 x y^{2}}{x^{2}+y^{2}} + i\ \frac{y^{3} -3 x^{2} y}{x^{2}+y^{2}}$ (1)

Now compute separately the $\lim_{(x,y) \rightarrow (0,0)$ of the real and imaginary part of (1)...

Kind regards

$\chi$ $\sigma$

**tonio** · November 14th 2010, 07:11 PM

Originally Posted by mukmar

Find $\displaystyle\lim_{z \to 0} \frac{\bar{z}^2}{z}$

Attempt: I thought that this limit might not exist since if you let $z = x + iy$ , then when approaching from y-axis, the limit equals $iy$ and when approaching from the x-axis, the limit equals $x$ .

However the solution says that limit is actually 0. I'm not sure how to get this result.

Any suggestions?

Another idea: write z in polar form, $z=re^{i\phi}\Longrightarrow z\rightarrow 0\Longleftrightarrow r\rightarrow 0\,,\,\overline{z}=re^{-i\phi}$ , so:

$\displaystyle{\frac{\overline{z}^2}{z}=\frac{r}{e^ {i\phi}}\xrightarrow [r\to 0]{} 0}$

Tonio

**FernandoRevilla** · November 14th 2010, 09:58 PM

Originally Posted by mukmar

Is there a limit law that says that the limit of the modulus is related to the limit of the function in any way?

Better:

$|\bar{z}^2/z-0|<\epsilon\Leftrightarrow |z-0|<\epsilon\quad (z\neq 0)$

Now, choose $\delta=\epsilon$ .

Regards.

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