complex number loci

leosco1995

Hey,

I need help with the last part. For the first inequality, I know that a vertical line is drawn through z = 1 but I don't know whether the right or left part of the line satisfies the inequality.

I don't understand the 2nd inequality. The argument of u is 153.4° but how do you show this inequality?

Help would really be appreciated. :smile:

loci.png47.3KB

Reply 1

BabyMaths

Original post by leosco1995

|z-0| < |z-2|

z is closer to 0 than it is to 2.

No time for the rest of your question just now. :tongue:

Reply 2

atsruser

Original post by leosco1995

The complex number

z-u

can be thought of as the vector from

u

z

i.e. as an arrow between those points. So:

0< \arg(z-u) < \frac{\pi}{4}

implies that the argument (i.e. angle with the positive x-axis) of this arrow is greater than 0 but less than 45 degrees.

Here,

z

can be chosen freely. Where can

z

be, so that the inequality above is satisfied?

Reply 3

leosco1995

Thanks for the quick responses guys. :smile:

Original post by atsruser

The complex number

z-u

can be thought of as the vector from

u

z

i.e. as an arrow between those points. So:

0< \arg(z-u) < \frac{\pi}{4}

implies that the argument (i.e. angle with the positive x-axis) of this arrow is greater than 0 but less than 45 degrees.

Here,

z

can be chosen freely. Where can

z

be, so that the inequality above is satisfied?

Edit - Nevermind. Is this diagram right (the diagonal is the first quadrant being 45 degrees)? If so then

z

can be anywhere from 0 to 45 degrees (from the +ve x-axis) right?

(edited 10 years ago)

Reply 4

atsruser

Original post by leosco1995

Thanks for the quick responses guys. :smile:

Edit - Nevermind. Is this diagram right (the diagonal is the first quadrant being 45 degrees)? If so then

z

can be anywhere from 0 to 45 degrees (from the +ve x-axis) right?

No, that's not right.

1. Draw a point representing

u

in the complex plane

2. Draw an arbitrary point

z

such that

\arg(z-u) = \frac{\pi}{4}

i.e. that the vector

z-u

lies at 45 degrees to the positive x-axis.

3. What other vectors will also have the same argument?

4. Which vectors emanating from

u

will have argument > 45 degrees?

5. Which vectors emanating from

u

will have argument < 45 degrees?

6. Which vectors emanating from

u

will have argument < 0 degrees?

7. So where can

z

be, so that the inequality is satisfied?

Reply 5

leosco1995

I'm bad. I tried again..

I guess if it's still wrong, then I have problems with understanding the z - u vector.

Reply 6

atsruser

Original post by leosco1995

I'm bad. I tried again..

I guess if it's still wrong, then I have problems with understanding the z - u vector.

z-u

is the arrow from

u

z

. That means that you *must* have an arrow emerging from

u

going to

z

. You seem to have drawn an arrow representing the complex number

u

, though it's hard to tell from your diagram.

If

\arg(z-u) < 45^{\circ}

, then that arrow cannot be steeper than (or as steep as) 45 degrees.

Have you followed the steps that I suggested above?

If so, put up an image showing your working.
If not, follow the steps that I suggested above, then show us your working.

Reply 7

leosco1995

I didn't quite understand your 2nd point in the instructions you gave.. I think that is where my problem lies actually. What is z? Any point in the graph (I feel so silly asking these questions so late)? The grey area in my diagram was the region I thought satisfied both inequalities but I'm not sure.

Reply 8

atsruser

Original post by leosco1995

I didn't quite understand your 2nd point in the instructions you gave.. I think that is where my problem lies actually. What is z? Any point in the graph

Yes. You can think of

z

as an arbitrary "test" point in the complex plane. If the line from

u

z

makes an angle with the real axis of between 0 and 45 degrees, then we let it into the select set of complex numbers that satisfy our requirement.

For example, a

z

that is directly "above"

u

would make an angle of 90 degrees with the real axis, so it wouldn't be in the set we want; a

z

that is directly "right" of

u

would make an angle of 0 degrees with the real axis, so that also wouldn't be in our set.

Does that help?

(I feel so silly asking these questions so late)? The grey area in my diagram was the region I thought satisfied both inequalities but I'm not sure.

The region of the plane that you want, by the way, is infinite in extent; your shading indicates a finite region.

Don't worry about asking questions; it's the only way to figure stuff out. (though sometimes you should be both asking *and* answering them)