Sketch the following sets in the complex plane

**terrorsquid** · September 7th 2011, 05:32 AM

I'm trying to sketch the following sets in the complex plane:

$a)~Re\left(\frac{1}{z}\right) < 1$

$b)~ |z+i|\ge|z-i|$

For $a)$ I found the real part of $\frac{1}{z}$ by setting $z = x + iy$ and got:

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

I then set this = 1 to try and graph something from which I can determine the values that are < 1:

$\frac{x}{x^2+y^2}=1$

$x = x^2+y^2$

$y = \sqrt{x-x^2}$

So I would have the top half of a circle starting at (0,0) and ending at (0,1), so my set would be drawn to represent [0,1) right?

The answer I have has a circle though with the equation $|y| \ge \sqrt{x-x^2}$ and I'm not sure how this point was achieved?

b) I'm not sure how to proceed since $z$ is abitrary but $i$ is defined and I can plot it at (0,1), so I don't see how I can compare them.

Thanks

**Prove It** · September 7th 2011, 05:38 AM

Originally Posted by terrorsquid

I'm trying to sketch the following sets in the complex plane:

$a)~Re\left(\frac{1}{z}\right) < 1$

$b)~ |z+i|\ge|z-i|$

For $a)$ I found the real part of $\frac{1}{z}$ by setting $z = x + iy$ and got:

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

I then set this = 1 to try and graph something from which I can determine the values that are < 1:

$\frac{x}{x^2+y^2}=1$

$x = x^2+y^2$

$y = \sqrt{x-x^2}$

So I would have the top half of a circle starting at (0,0) and ending at (0,1), so my set would be drawn to represent [0,1) right?

The answer I have has a circle though with the equation $|y| \ge \sqrt{x-x^2}$ and I'm not sure how this point was achieved?

b) I'm not sure how to proceed since $z$ is abitrary but $i$ is defined and I can plot it at (0,1), so I don't see how I can compare them.

Thanks

You want

$\displaystyle \begin{align*} \frac{x}{x^2 + y^2} &< 1 \\ x &< x^2 + y^2 \\ x - x^2 &< y^2 \\ -\left(x^2 - x\right) &< y^2 \\ -\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2\right] &< y^2 \\ -\left[\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right] &< y^2 \\ -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4} &< y^2 \\ \frac{1}{4} &< \left(x - \frac{1}{2}\right)^2 + y^2 \end{align*}$

So it's everything inside, but not including, the circle of radius $\displaystyle \frac{1}{2}$ and centred at $\displaystyle (x, y) = \left(\frac{1}{2}, 0\right)$

**Prove It** · September 7th 2011, 05:43 AM

Originally Posted by terrorsquid

I'm trying to sketch the following sets in the complex plane:

$a)~Re\left(\frac{1}{z}\right) < 1$

$b)~ |z+i|\ge|z-i|$

For $a)$ I found the real part of $\frac{1}{z}$ by setting $z = x + iy$ and got:

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

I then set this = 1 to try and graph something from which I can determine the values that are < 1:

$\frac{x}{x^2+y^2}=1$

$x = x^2+y^2$

$y = \sqrt{x-x^2}$

So I would have the top half of a circle starting at (0,0) and ending at (0,1), so my set would be drawn to represent [0,1) right?

The answer I have has a circle though with the equation $|y| \ge \sqrt{x-x^2}$ and I'm not sure how this point was achieved?

b) I'm not sure how to proceed since $z$ is abitrary but $i$ is defined and I can plot it at (0,1), so I don't see how I can compare them.

Thanks

b)

$\displaystyle \begin{align*} |z + i| &\geq |z - i| \\ |x + iy + i| &\geq |x + iy - i| \\ |x + (y + 1)i| &\geq |x + (y - 1)i| \\ \sqrt{x^2 + (y + 1)^2} &\geq \sqrt{x^2 + (y - 1)^2} \\ x^2 + (y + 1)^2 &\geq x^2 + (y - 1)^2 \\ (y + 1)^2 &\geq (y - 1)^2 \\ y^2 + 2y + 1 &\geq y^2 - 2y + 1 \\ 2y &\geq -2y \\ 4y &\geq 0 \\ y &\geq 0 \end{align*}$

So the required region is everything above and including the $\displaystyle x$ axis.

**terrorsquid** · September 7th 2011, 05:58 AM

Wow, thanks!

a) I can see what you have done but it seems so convoluted; how would I have known to do that and why did you choose $-\frac{1}{2}$ (did you choose it? or work it out somehow?)

b) I didn't understand the transition from line 3-4 (the jump from |...| to \sqrt...). Everythign leading up to it and following it is fine, just that step.

**Prove It** · September 7th 2011, 06:14 AM

Originally Posted by terrorsquid

Wow, thanks!

a) I can see what you have done but it seems so convoluted; how would I have known to do that and why did you choose $-\frac{1}{2}$ (did you choose it? or work it out somehow?)

b) I didn't understand the transition from line 3-4 (the jump from |...| to \sqrt...). Everythign leading up to it and following it is fine, just that step.

a) Completing the Square, enables you to put the circle into standard form.

b) Surely you know how to evaluate the modulus of a complex number... $\displaystyle |a + bi| = \sqrt{a^2 + b^2}$ .

**terrorsquid** · September 7th 2011, 06:25 AM

Originally Posted by Prove It

a) Completing the Square, enables you to put the circle into standard form.

b) Surely you know how to evaluate the modulus of a complex number... $\displaystyle |a + bi| = \sqrt{a^2 + b^2}$ .

Ah, of course! to both

Thanks a lot [SOLVED]

**Plato** · September 7th 2011, 06:31 AM

Originally Posted by terrorsquid

$b)~ |z+i|\ge|z-i|$

Here is a different way to approach this question.
$|z-w|$ is just the distance between $z~\&~w$ .
So you want the complex numbers whose distance from $\mathif{-i}$ is greater than or equal to the distance to $\mathif{i}$ .

**terrorsquid** · September 7th 2011, 06:45 AM

Originally Posted by Plato

Here is a different way to approach this question.
$|z-w|$ is just the distance between $z~\&~w$ .
So you want the complex numbers whose distance from $\mathif{-i}$ is greater than or equal to the distance to $\mathif{i}$ .

Ah yeh, this is the method that someone mentioned to me but they couldn't seem to explain my question. Shouldn't it be everything below and including the $x$ axis if you want numbers whos distance between $z$ and $+i$ is $\ge$ the distance between $z$ and $-i$ , i.e. all the numbers that are closer to $-i$ (lesser or equal distance)? I assume I am misunderstanding something :S

**Plato** · September 7th 2011, 06:54 AM

Originally Posted by terrorsquid

Ah yeh, this is the method that someone mentioned to me but they couldn't seem to explain my question. Shouldn't it be everything below and including the $x$ axis if you want numbers whos distance between $z$ and $+i$ is $\ge$ the distance between $z$ and $-i$ , i.e. all the numbers that are closer to $-i$ (lesser or equal distance)? I assume I am misunderstanding something :S

No $|z\mathif{-i}|$ is the distance from $z\text{ to }\mathif{i}$
While $|z+\mathif{i}|$ is the distance from $z\text{ to }\mathif{-i}$ .
So it is the closed upper half-plane.

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