What is the limit of the function as z approaches 0?

[Benzoate]

Homework Statement

Used the definition of a limit to prove that as z=>0 lim (z bar)^2/(z)=0

Homework Equations

abs(f(z)-w(0)) < eplison whenever abs(z-z(0)) < lower case delta

The Attempt at a Solution

let z=x+iy and z bar = x-iy

z=(x,y)

Since limit of function is approaches origin, there are two cases when the limit approaches the origin: when (x,0) and when (y,0)

first case(real axis): z=>(x,0) lim (z bar)^2/(z)=0 => (x-i*0)^2/(x+i*0)=(x^2)/x= x

second case(imaginary axis) : z=>(0,y) lim (z bar)^2/(z)=0 => (0-i*y)^2/(0+i*y)=(i*y)^2/(i*y)= i*y

Both cases each time show that the function has two different limits, and that the limit of the function does not approach zero in either case.

So how can the limit of the function be zero?

 

[morphism]

I can't understand what you're doing. What does "when (x,0) and when (y,0)" mean?

Also, you should have $0 < |z - z_0| < \delta$ in the definition of a limit; the "0 <" part is important.

Now to solve your problem, I recommend that you first prove the following basic facts:
(1) |z/w| = |z|/|w| (w$\neq$0)
(2) |$\overline{z}$| = |z|.

 

[Benzoate]

I can't understand what you're doing. What does "when (x,0) and when (y,0)" mean?

Also, you should have $0 < |z - z_0| < \delta$ in the definition of a limit; the "0 <" part is important.

Now to solve your problem, I recommend that you first prove the following basic facts:
(1) |z/w| = |z|/|w| (w$\neq$0)
(2) |$\overline{z}$| = |z|.

So I have to show that |z_bar^2/z|=|z_bar*z_bar/z|=|z_bar|*|z_bar|/z

 

[morphism]

I was hoping you'd see that, for z$\neq$0,

$$\left| \frac{\bar{z}^2}{z} \right| = |z|.$$

 

[Benzoate]

I was hoping you'd see that, for z$\neq$0,

$$\left| \frac{\bar{z}^2}{z} \right| = |z|.$$

but wouldn't |z_bar^2/z| => |(x-iy)^2/(x+iy)| since z_bar is the conjugate of z. So how can |z_bar^2/z|=|z|?

 

[HallsofIvy]

Homework Statement

Used the definition of a limit to prove that as z=>0 lim (z bar)^2/(z)=0

Homework Equations

abs(f(z)-w(0)) < eplison whenever abs(z-z(0)) < lower case delta

The Attempt at a Solution

let z=x+iy and z bar = x-iy

z=(x,y)

Since limit of function is approaches origin, there are two cases when the limit approaches the origin: when (x,0) and when (y,0)

first case(real axis): z=>(x,0) lim (z bar)^2/(z)=0 => (x-i*0)^2/(x+i*0)=(x^2)/x= x

second case(imaginary axis) : z=>(0,y) lim (z bar)^2/(z)=0 => (0-i*y)^2/(0+i*y)=(i*y)^2/(i*y)= i*y

Both cases each time show that the function has two different limits, and that the limit of the function does not approach zero in either case.

So how can the limit of the function be zero?

?? If z is approaching 0, both x and y go to 0. Those certainly do go to zero!

 

[complexnumber]

please see attachment

 

[complexnumber]

$$\left\vert \frac{\bar{z}^2}{z} - 0 \right\vert = \left\vert \frac{\bar{z}^2}{z} \right\vert = |\bar{z} \bar{z} z^{-1}| = |\bar{z}| |\bar{z}| |z^{-1}| = |z||z||z^{-1}| = |z||z z^{-1}| = |z| < \varepsilon$$