Show that the given function is continuous

[chwala]

Homework Statement

Show that the function $f :\mathbb{C} → \mathbb{C}$ defined by $f(z) = {\sqrt 3} i^2 |z|$ is a continous function.

Relevant Equations

complex numbers.

Refreshing... going through the literature i may need your indulgence or direction where required. ...of course i am still studying on the proofs of continuity...the limits and epsilons... in reference to continuity of functions...

From my reading, A complex valued function is continous if and only if both its real part and imaginary part are continous.
Firstly
$f(z)= x+iy = -\sqrt 3 x^2 -\sqrt 3 y^2 i=u(x,y)+iv(x,y)$,that is from the concept under Cauchy-Riemann equations. Therefore, $u(x,y)=-\sqrt 3 x^2$ and $v(x,y)=-\sqrt 3 y^2$ of which both $u(x,y)$ and $v(x,y)$ are continous.

 

[Hill]

How did you get this:
$-\sqrt 3 x^2 -\sqrt 3 y^2 i$
?

 

[chwala]

How did you get this:
$-\sqrt 3 x^2 -\sqrt 3 y^2 i$
?

$|z| = |x+iy| = x^2+y^2$ unless i am missing something...

 

[Hill]

$|z| = |x+iy| = x^2+y^2$ unless i am missing something...

It does not answer my question AND this is incorrect.

 

[chwala]

It does not answer my question AND this is incorrect.

okay...am back to the drawing board...later. I am missing the square root sign...typo...let me take a break.

$\sqrt{x^2 + y^2}$

 

[chwala]

It does not answer my question AND this is incorrect.

From where i left, i have

$f(z)=\sqrt 3 ⋅ i^2|z|=-\sqrt 3 |x+iy| = -\sqrt 3 {\sqrt{x^2 + y^2}}$

We know that $z=x+iy$ therefore using the cauchy Riemann equations we shall
let $u(x,y) =x$ and $v(x,y)=iy$
then,

$u_x= -\dfrac{\sqrt{3}}{2} ⋅(x^2+y^2)^{-0.5}⋅ 2x= -\dfrac{2x\sqrt{3}}{2\sqrt{x^2+y^2}} =-\dfrac{\sqrt{3}x}{\sqrt{x^2+y^2}}$ and

$u_y= -\dfrac{\sqrt{3}}{2}⋅ (x^2+y^2)^{-0.5} ⋅2y=-\dfrac{2y\sqrt{3}}{2\sqrt{x^2+y^2}}=-\dfrac{\sqrt{3}y}{\sqrt{x^2+y^2}}$

$u(x,y)$ is continous and $v(x,y)$ is continous and therefore $f(z)$ will be continous.

 

[Hill]

let $u(x,y) =x$ and $v(x,y)=iy$

I don't understand what it means and don't see you using it anywhere.

 

[pasmith]

Homework Statement: Show that the function $f :\mathbb{C} → \mathbb{C}$ defined by $f(z) = \sqrt 3 i^2 |z|$ is a continous function.
Relevant Equations: complex numbers.

Use the definition of continuity directly. $f : \mathbb{C} \to \mathbb{C}$ is continuous at $w \in \mathbb{C}$ iff for every $\epsilon > 0$ there exists $\delta > 0$ such that for every $\mathbb{z} \in \mathbb{C}$, if $|z - w| < \delta$ then $|f(z) - f(w)| < \epsilon$. Consider that $$|z| = |z - w + w| \leq |z - w| + |w|$$ and $$|w| = |w - z + z| \leq |w - z| + |z|.$$ Hence $$||z| - |w|| \leq |z - w|.$$

 

[chwala]

Use the definition of continuity directly. $f : \mathbb{C} \to \mathbb{C}$ is continuous at $w \in \mathbb{C}$ iff for every $\epsilon > 0$ there exists $\delta > 0$ such that for every $\mathbb{z} \in \mathbb{C}$, if $|z - w| < \delta$ then $|f(z) - f(w)| < \epsilon$. Consider that $$|z| = |z - w + w| \leq |z - w| + |w|$$ and $$|w| = |w - z + z| \leq |w - z| + |z|.$$ Hence $$||z| - |w|| \leq |z - w|.$$

Can I let $w=0$. This is an area I need to study...and are you implying one cannot use the Cauchy Riemann equations?

 

[Orodruin]

Can I let $w=0$. This is an area I need to study...and are you implying one cannot use the Cauchy Riemann equations?

No. The only requirement on w is that $|z-w|<\delta$.it should be clise to whatever z you want to show continuity at.

Alternatively just show that the preimage of any open set is an open set. That should be pretty straightforward.

 

[nuuskur]

Alternatively, we see that $f$ is a composition of polynomial and modulus, both of which are continuous.

Cauchy-Riemann stuff is relevant for complex differentiability.

 

[Mark44]

Show that the function $f :\mathbb{C} → \mathbb{C}$ defined by $f(z) = \sqrt 3 i^2 |z|$ is a continous function.

You don't have braces around the square root radicand.
Do you mean this -- $\sqrt 3 \cdot i^2 |z|$?
Or this -- $\sqrt{3 i^2}|z|$?

The first would simplify to $-\sqrt 3 |z|$ and the second would simplify to $\sqrt{-3}|z| = i\sqrt 3 |z|$.

It should be very simple to show continuity for either one.

Regarding TeX, keep in mind that if what comes after an operator is a single character, you can omit braces, but if what follows is two or more characters, braces are necessary.

For example, I can omit braces to write $x^2$ (unrendered this is ##x^2##), but braces are needed for $x^{-1}$ (unrendered: ##x^{-1}##). Without braces, this comes out as $x^-1$; i.e., the 1 doesn't get rendered as a superscript.

 

[WWGD]

If your $i^2$ refers to the Complex number $i$, then your map is actually $\mathbb R$- valued. As such, it's a product of continuous functions.

 

[chwala]

If your $i^2$ refers to the Complex number $i$, then your map is actually $\mathbb R$- valued. As such, it's a product of continuous functions.

yes, the $i$ refers to the complex number.

 

[chwala]

You don't have braces around the square root radicand.
Do you mean this -- $\sqrt 3 \cdot i^2 |z|$?
Or this -- $\sqrt{3 i^2}|z|$?

The first would simplify to $-\sqrt 3 |z|$ and the second would simplify to $\sqrt{-3}|z| = i\sqrt 3 |z|$.

It should be very simple to show continuity for either one.

Regarding TeX, keep in mind that if what comes after an operator is a single character, you can omit braces, but if what follows is two or more characters, braces are necessary.

For example, I can omit braces to write $x^2$ (unrendered this is ##x^2##), but braces are needed for $x^{-1}$ (unrendered: ##x^{-1}##). Without braces, this comes out as $x^-1$; i.e., the 1 doesn't get rendered as a superscript.

The first one...

 

[Mark44]

The first one...

Then you want to write it as $\sqrt{3 i^2}|z| = \sqrt{-3}|z| = i\sqrt 3 |z|$. I assume you know how to right click on the above to show the TeX that I wrote.

Edit: I mistook what the first one was. That one should have been written as $-\sqrt 3 i^2|z| = -\sqrt 3 |z|$.

 

[docnet]

My friends, I

 

[Orodruin]

The first one was this:

Do you mean this -- $\sqrt 3 \cdot i^2 |z|$?

 

[chwala]

The first one was this:

Yes @Orodruin , sorry I was unable initially to put parenthesis...but @Mark44 advised how to do it .. I didn't want to amend original post following the threads...

 

[Mark44]

The first one was this

Fixed in my post... Thanks for pointing this out!

 

[mathwonk]

forgive me if I misunderstand the question. either you want to know how to give a formal proof of continuity, or to understand why it is true. In my opinion, if you understand why, you have a better chance of giving a proof, so I address that.

Continuity means that whenever the inputs are close, the outputs are also close.

'first consider the function |z|. Note that |z| is the distance from z to the origin. thus one must show that if z is close to w, then z is almost as close to the origin as w. This is called the triangle inequality.

I.e. the distance from z to the origin is less than the sum of the distance from w to the origin plus the distance from z to w. Thus if z is near w, then z is almost as close to the origin as w.

 

[chwala]

forgive me if I misunderstand the question. either you want to know how to give a formal proof of continuity, or to understand why it is true. In my opinion, if you understand why, you have a better chance of giving a proof, so I address that.

Continuity means that whenever the inputs are close, the outputs are also close.

'first consider the function |z|. Note that |z| is the distance from z to the origin. thus one must show that if z is close to w, then z is almost as close to the origin as w. This is called the triangle inequality.

I.e. the distance from z to the origin is less than the sum of the distance from w to the origin plus the distance from z to w. Thus if z is near w, then z is almost as close to the origin as w.

The question is a past exam question -Analysis paper. The question is written as it is on the paper.

 

[mathwonk]

forgive me, I assumed you wanted to understand it. that is what I am addressing.

 

[chwala]

forgive me if I misunderstand the question. either you want to know how to give a formal proof of continuity, or to understand why it is true. In my opinion, if you understand why, you have a better chance of giving a proof, so I address that.

Continuity means that whenever the inputs are close, the outputs are also close.

'first consider the function |z|. Note that |z| is the distance from z to the origin. thus one must show that if z is close to w, then z is almost as close to the origin as w. This is called the triangle inequality.

I.e. the distance from z to the origin is less than the sum of the distance from w to the origin plus the distance from z to w. Thus if z is near w, then z is almost as close to the origin as w.

The question is a past exam question -Analysis paper. The question is written as it is on the paper.

forgive me, I assumed you wanted to understand it. that is what I am addressing.

I really appreciate it @mathwonk ...my undergraduate professors rushed on the chapters to do with Analysis plus I didn't put much work afterwards..I just have to self study... Cheers man.

 

[martinbn]

This is a very strange question! Clearly the constant $\sqrt{3}i^2$ plays no role, why is it discussed so much! So the question is to show that $f(z)=|z|$ is continuous. Isn't this obvious given what continuous means! May be you can start with the definition of continuous you use.

 

[WWGD]

If your function is $f(z)=|z|$, then you can check whether the inverse image of an open set is open. It will depend on your domain, range. In this case, $|z|$ is Real-valued. An open interval$(a,b)$ would have as an inverse image all circles of radius $r; a<r<b$.

 

[Orodruin]

If your function is $f(z)=|z|$, then you can check whether the inverse image of an open set is open. It will depend on your domain, range. In this case, $|z|$ is Real-valued. An open interval$(a,b)$ would have as an inverse image all circles of radius $r; a<r<b$.

I tried to hint at this, but OP didn't seem to bite:

Alternatively just show that the preimage of any open set is an open set. That should be pretty straightforward.

It should be added that the shape of the preimage is called an annulus.

 

[chwala]

I tried to hint at this, but OP didn't seem to bite:


It should be added that the shape of the preimage is called an annulus.

I am following the steps slowly ...