# Thread: limit of complex variable

1. ## limit of complex variable

lim z->1
[z(conjugate z) + 2z - (conjugate z) -2 ]
---------------------------------------
z^2 - 1

answer given is z / 2

but i stuck at

2+(conjugate z)
----------------
z + 1

can anyone help pls...
thank you...

2. Originally Posted by kennytsh
lim z->1
[z(conjugate z) + 2z - (conjugate z) -2 ]
---------------------------------------
z^2 - 1

answer given is z / 2

but i stuck at

2+(conjugate z)
----------------
z + 1

can anyone help pls...
thank you...
It can be written:

$\displaystyle \frac{z\bar{z} - \bar{z}}{(z+1)(z-1)} + \frac{2z - 2}{(z+1)(z-1)}$

$\displaystyle \frac{\bar{z}(z-1)}{(z+1)(z-1)} + \frac{2(z-1)}{(z+1)(z-1)}$

$\displaystyle \frac{\bar{z}}{z+1} + \frac{2}{z+1} = \frac{\bar{z} + 2}{z+1}$

Now multiply top and bottom by z:

$\displaystyle \frac{\bar{z}z + 2z}{z^2+z}$

And remember that $\displaystyle \bar{z}z = |z|^2$

PS: The answer can't be $\displaystyle \frac{z}{2}$...

3. Hello, kennytsh!

$\displaystyle \lim_{z\to1} \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1}$

Answer given: $\displaystyle \frac{z}{2}$ . . . . I don't agree

but stuck at: .$\displaystyle \frac{\overline{z}+2}{z+1}\quad {\color{blue}\leftarrow\:\text{Right!}}$

We have: .$\displaystyle \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1} \;=\;\frac{z(\overline{z}+2) - (\overline{z}+2)}{z^2-1}\;=\;\frac{(z-1)(\overline{z}+2)}{(z-1)(z+1)} \;=\;\frac{\overline{z}+2}{z+1}$

Now we have: .$\displaystyle \lim_{z\to1}\frac{\overline{z}+2}{z+1} \;=\;\lim_{z\to1} \frac{a-bi + 2}{a + bi + 1}$

But if $\displaystyle z\to1$, then: .$\displaystyle \begin{array}{c}a \to 1 \\ b\to 0 \end{array}$

. . So we have: .$\displaystyle \lim_{\begin{array}{c}^{a\to1}\\ [-2mm]^{b\to0}\end{array}} \frac{a - bi + 2}{a + bi + 1} \;=\;\frac{1 - 0 + 2}{1 + 0+ 1} \;=\;\frac{3}{2}$

4. Originally Posted by Soroban
Hello, kennytsh!

We have: .$\displaystyle \frac{z\overline{z} + 2z - \overline{z} - 2}{z^2-1} \;=\;\frac{z(\overline{z}+2) - (\overline{z}+2)}{z^2-1}\;=\;\frac{(z-1)(\overline{z}+2)}{(z-1)(z+1)} \;=\;\frac{\overline{z}+2}{z+1}$

Now we have: .$\displaystyle \lim_{z\to1}\frac{\overline{z}+2}{z+1} \;=\;\lim_{z\to1} \frac{a-bi + 2}{a + bi + 1}$

But if $\displaystyle z\to1$, then: .$\displaystyle \begin{array}{c}a \to 1 \\ b\to 0 \end{array}$

. . So we have: .$\displaystyle \lim_{\begin{array}{c}^{a\to1}\\ [-2mm]^{b\to0}\end{array}} \frac{a - bi + 2}{a + bi + 1} \;=\;\frac{1 - 0 + 2}{1 + 0+ 1} \;=\;\frac{3}{2}$

Surely it's simply to state that $\displaystyle \bar{z}z = |z|^2$, evaluate the limit of just z.