Complex Variables Question (should be easy)

[DEMJ]

Homework Statement

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

Homework Equations

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

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.

 

[gabbagabbahey]

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

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

 

[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.

Where should I start with this equation:
$$Re(\bar{z} - i) = 2$$

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?

 

[gabbagabbahey]

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.

Yup!

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

So $$Re(\bar{z} - i) = Re(\bar{z}) -Re(i) = x + -Re(i)$$

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

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

right?

 

[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?

 

[gabbagabbahey]

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.

Looks good to me!

Is there any way to rate people who reply?

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.