# Complex Variables Question (should be easy)

1. Aug 30, 2009

### DEMJ

1. The problem statement, all variables and given/known data

Sketch the set of points determined $$\mid 2\bar z + i \mid = 4$$.

2. Relevant equations

$$\mid z - z_0 \mid = r$$ and $$\mid \bar z \mid = \mid z \mid$$

3. The attempt at a solution
I know that it will be the circle with radius = 4 and $$z_0$$ is the center of the circle and that the points that are relevant are the ones on the circle only because it is an equals sign. So with this equation what is throwing me off is the $$2z$$. What I did was (not sure if this is part is legal) is divide both sides by 2 and obtained $$\mid z - (-\frac{1}{2}i) \mid = 2$$.

So this means I have a circle of $$r=2$$ and $$z_0 = (0, -\frac{1}{2})$$ to draw. Please let me know if this is correct or not. Thank you.

2. Aug 30, 2009

### gabbagabbahey

Dividing by 2 is perfectly fine (since for any any real number $c$, you know $|cz|=c|z|$), but you have another mistake:

$$|\bar{z}|=|z|\implies |2\bar{z}+i|=|\overline{(2\bar{z}+i)}|=|2z-i|\neq|2z+i|$$

3. Aug 30, 2009

### DEMJ

Oh so I should have just took the conjugate of $$|2\bar{z} + i| = |\overline{2\bar{z} + i}| = |2z -i|$$ then you get $$|z - \frac{1}{2}i| = 2$$

So I should draw my circle with a center at $$(0,\frac{1}{2})$$ and radius 2.

Thank you very much for the help I believe this solves that question.

$$Re(\bar{z} - i) = 2$$

I know that:
$$Re(\bar{z}) = Re(z) = \frac{z+\bar{z}}{2} = x$$

So $$Re(\bar{z} - i) = Re(\bar{z}) -Re(i) = x + -Re(i)$$ Then the only thing I can think of is since i = (0,1) then the R(i) = (0,0). How does this look?

4. Aug 30, 2009

### gabbagabbahey

Yup!

No, if $Re(z) = \frac{z+\bar{z}}{2}$, then

$$Re(\bar{z} - i)=\frac{(\bar{z} - i)+\overline{(\bar{z} - i)}}{2}$$

right?

5. Aug 30, 2009

### DEMJ

Yes you are right. So by this we have $$Re(\bar{z} - i)=\frac{(\bar{z} - i)+\overline{(\bar{z} - i)}}{2} = \frac{(\bar{z} - i) + (z + i)}{2} = \frac{(x - iy - i) + (x + iy + i)}{2} = \frac{2x}{2} = x$$

So it's just the points that lie on the line of x = 2. I am pretty sure this is right and I am thankful for your great responses. Is there any way to rate people who reply?

6. Aug 30, 2009

### gabbagabbahey

Looks good to me!

I think there is a poll held in the general discussion forum at the end of the year for homework helper of the year, but other than that I don't think so.