Continuity of a complex-valued function

**deepakpc007** · April 28th 2011, 08:09 AM

Define the function f(z)=z*Real Part(z)/|z|,z!= 0
0 ,z=0
Prove that f(z) is continuous in the entire complex plane.

Pls somebody help me with dis one......thanx in advance.

**Plato** · April 28th 2011, 08:26 AM

Originally Posted by deepakpc007

Define the function f(z)=z*Real Part(z)/|z|,z!= 0
0 ,z=0
Prove that f(z) is continuous in the entire complex plane.

If $z=re^{\mathif{i}\theta}$ then $\frac{z\cdot\text{Re}(z)}{|z|}=\frac{(re^{\mathif{ i}\theta})(r\cos(\theta)}{r}$ .

Now what happens when $r\to 0~?$

**deepakpc007** · April 28th 2011, 08:51 AM

Originally Posted by Plato

If $z=re^{\mathif{i}\theta}$ then $\frac{z\cdot\text{Re}(z)}{|z|}=\frac{(re^{\mathif{ i}\theta})(r\cos(\theta)}{r}$ .

Now what happens when $r\to 0~?$

when $r\to 0~?$ , $z=re^{\mathif{i}\theta}$ then $\frac{z\cdot\text{Re}(z)}{|z|}=\frac{(re^{\mathif{ i}\theta})(r\cos(\theta)}{r}$ =infinity???I think so.

**Plato** · April 28th 2011, 08:58 AM

Originally Posted by deepakpc007

when $r\to 0~?$ , $z=re^{\mathif{i}\theta}$ then $\frac{z\cdot\text{Re}(z)}{|z|}=\frac{(re^{\mathif{ i}\theta})(r\cos(\theta)}{r}$ =infinity???I think so.

Oh my no.
Is this true: $r\not= 0$ so $\frac{(re^{\mathif{i}\theta})(r\cos(\theta)}{r}=(r e^{\mathif{i}\theta})(\cos(\theta))~?$

**deepakpc007** · April 28th 2011, 09:02 AM

Originally Posted by Plato

Oh my no.
Is this true: $r\not= 0$ so $\frac{(re^{\mathif{i}\theta})(r\cos(\theta)}{r}=(r e^{\mathif{i}\theta})(\cos(\theta))~?$

Yes thats true.How can i prove with this?

**Plato** · April 28th 2011, 09:12 AM

Originally Posted by deepakpc007

Yes thats true.How can i prove with this?

What is there to prove.
As $r\to 0$ the other factors are bounded, so what is the limit?

**deepakpc007** · April 28th 2011, 09:23 AM

Originally Posted by Plato

What is there to prove.
As $r\to 0$ the other factors are bounded, so what is the limit?

Sorry sir,i dont know.i am not dat good in maths.please help me.

Edit : As $r\to 0$ the other factors are bounded,means a finite limit.Is it so?
Limit is 0 and hence the function is continuous in the entire complex plane.Am i right?

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