You are Here: Home >< Maths

# Argument watch

1. Use an Argand diagram to find, in the form , the complex number which satisfies the following pair of equations:

and

I've tried to illustrate these equations on a diagram but I'm struggling to find the complex number which satisfies the equations.
2. (Original post by Femto)
Use an Argand diagram to find, in the form , the complex number which satisfies the following pair of equations:

and

I've tried to illustrate these equations on a diagram but I'm struggling to find the complex number which satisfies the equations.
What diagram did you get?
3. One line from the origin at angle 7/12 pi from the x axis and another line from the point 2 + 2i at an angle of 11/12 pi from the horizontal to that point.
4. (Original post by Femto)
One line from the origin at angle 7/12 pi from the x axis and another line from the point 2 + 2i at an angle of 11/12 pi from the horizontal to that point.
So, you end up with a rather special triangle.

What's the length of a side?

What's the angle between the point and the positive real axis?

Hence what's the point?
5. Sorry for the late reply; thanks for your help I managed to tackle it in the end!
6. I have another question:

A complex number satisfies . Find:

(a) the greatest value of

I've solved this part, the answer is .

(b) The difference between the greatest and least values of

This part I'm stuck on. Can anyone help please?
7. Surely it is

This is not the answer in the back of the book..
8. (Original post by Femto)
I have another question:

A complex number satisfies . Find:

(a) the greatest value of

I've solved this part, the answer is .

(b) The difference between the greatest and least values of

This part I'm stuck on. Can anyone help please?
Draw a diagram.

Forget about the complex plane, just think in terms of normal coordinate geometry.

The modulus equation is a circle centre (3,4) radius 2 as you have clearly already realised.

The least and greatest values of arg z will be tangents to the circle starting at (0,0). Bit of trig (remember radius and tangent meet at 90 degs) and you are home.
9. (Original post by Mr M)
Draw a diagram.

Forget about the complex plane, just think in terms of normal coordinate geometry.

The modulus equation is a circle centre (3,4) radius 2 as you have clearly already realised.

The least and greatest values of arg z will be tangents to the circle starting at (0,0). Bit of trig (remember radius and tangent meet at 90 degs) and you are home.
Hmm this is what I've done, although I haven't joined the line from the origin which goes to the circle to the centre of the circle - should I have done? Why would it make a difference to which angle I get if I join a line to the circle then straight down to the real axis?
10. (Original post by Femto)
Hmm this is what I've done, although I haven't joined the line from the origin which goes to the circle to the centre of the circle - should I have done? Why would it make a difference to which angle I get if I join a line to the circle then straight down to the real axis?
I see what you have done.

The point 3 + 2i (and 1 + 4i) is on the circle but this does not give you the max and min arguments. On your picture you can presumably see the line from the origin is not perpendicular to the radius in each case.
11. (Original post by Mr M)
Draw a diagram.

Forget about the complex plane, just think in terms of normal coordinate geometry.

The modulus equation is a circle centre (3,4) radius 2 as you have clearly already realised.

The least and greatest values of arg z will be tangents to the circle starting at (0,0). Bit of trig (remember radius and tangent meet at 90 degs) and you are home.
Ignore what I've just said.

I make the answer to be:

Which is nearly right but for some reason the back of the book has

Could you check my workings?

So I've got two lines heading to the circle, the first is about an angle from the real axis and is my smaller angle. I know that the distance to the centre of the circle from the origin is so surely:

Then I've got another angle, from the real axis which goes to the maximum angle which I make out to be:

So surely the difference is simply
12. (Original post by Femto)
Ignore what I've just said.

I make the answer to be:

Which is nearly right but for some reason the back of the book has

Could you check my workings?

So I've got two lines heading to the circle, the first is about an angle from the real axis and is my smaller angle. I know that the distance to the centre of the circle from the origin is so surely:

Then I've got another angle, from the real axis which goes to the maximum angle which I make out to be:

So surely the difference is simply
Wow you have made it incredibly complicated.

You don't need to work out the arguments. You just want the difference between them.

Draw a kite with a line segment of symmetry passing through (0, 0) and (3, 4) - the centre of the circle. The line segment has length 5. The two short sides of the kite have length 2. The kite has two 90 degrees angles.

The kite is symmetrical about the line segment. Call the angle INSIDE the kite at the origin 2x. Hence x = arcsin (2/5). What does 2x equal?
13. (Original post by Mr M)
Wow you have made it incredibly complicated.

You don't need to work out the arguments. You just want the difference between them.

Draw a kite with a line segment of symmetry passing through (0, 0) and (3, 4) - the centre of the circle. The line segment has length 5. The two short sides of the kite have length 2. The kite has two 90 degrees angles.

The kite is symmetrical about the line segment. Call the angle INSIDE the kite at the origin 2x. Hence x = arcsin (2/5). What does 2x equal?
Hmm, I thought it was the angle from the real axis though? Or is it the angle inside the kite which is the argument? I'm so confused
14. (Original post by Femto)
Hmm, I thought it was the angle from the real axis though? Or is it the angle inside the kite which is the argument? I'm so confused
The DIFFERENCE between the two arguments is the angle inside the kite and that is what you are being asked to find.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 20, 2011
Today on TSR

### Are you living on a tight budget at uni?

From budgets to cutbacks...

### University open days

1. Loughborough University
Fri, 21 Sep '18
2. University of Cambridge
Fri, 21 Sep '18
3. Richmond, The American International University in London
Fri, 21 Sep '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams