Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    5. (a) Sketch on an Argand diagram the circle C whose equation is
    | z - root3 - i | = 1
    (b) Mark the point P on C at which |z| is a minimum. Find this minimum value.
    (c) Mark the point Q on C at which arg z is a maximum. Find this maximum value.

    I've done part (a) and drawn a circle with centre (root3, 1).
    I don't know how to do part (b), surely this would be the the line from the origin to the circle with the shortest distance but how do I find it? Don't know how to do part (c) either.
    Offline

    16
    ReputationRep:
    |z| represents the distance from the origin. Your equation represents a circle. The max and min values of |z| are the distances from the origin of the points furthest/closest to the origin which are on that circle.

    Once you've drawn the circle, start by drawing a line from the origin to the centre of the circle and then mark on the distances you know.
    Offline

    1
    ReputationRep:
    (Original post by thers)
    5. (a) Sketch on an Argand diagram the circle C whose equation is
    | z - root3 - i | = 1
    (b) Mark the point P on C at which |z| is a minimum. Find this minimum value.
    (c) Mark the point Q on C at which arg z is a maximum. Find this maximum value.

    I've done part (a) and drawn a circle with centre (root3, 1).
    I don't know how to do part (b), surely this would be the the line from the origin to the circle with the shortest distance but how do I find it? Don't know how to do part (c) either.
    Finding the minimum would be by the use of Pythagoras' theorem, then the furthest distance would be shortest distance + diameter of circle.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    1
    ReputationRep:
    Ok I got |z| to be 1 but in the back of the textbook where the answers are it shows the diagram but P is on the other side of the circle, furthest away from the origin and says |z| is 3. :confused:
    Offline

    1
    ReputationRep:
    (Original post by thers)
    Ok I got |z| to be 1 but in the back of the textbook where the answers are it shows the diagram but P is on the other side of the circle, furthest away from the origin and says |z| is 3. :confused:
    |z - \sqrt{3} - i| = 1 represents a circle with centre (\sqrt{3}, 1) and a radius of 1.

    |z|_{min} = length of line from origin to centre of circle minus the radius.

    |z|_{max} = length of the line from origin to centre of circle add the radius.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Joshmeid)
    |z - \sqrt{3} - i| = 1 represents a circle with centre (\sqrt{3}, 1) and a radius of 1.

    |z|_{min} = length of line from origin to centre of circle minus the radius.

    |z|_{max} = length of the line from origin to centre of circle add the radius.
    so |z| 1 is correct?

    So how do I do part (c) now?
    I drew a line from the origin to the left of the circle. I created a right angled triangle with adjacent side root3 - 1 and opposite side 1 relative to the unknown angle. I get arctan(1/(root3 - 1)) but according to the textbook that is wrong - it should be pi/3.
    Offline

    1
    ReputationRep:
    b is correct.

    For C, Look at your drawing, we can get the size of the angle between the positive x-axis and the |z| as we have the lengths for them, the maximum of argz will therefore be double that.
    • Thread Starter
    Offline

    1
    ReputationRep:
    I don't understand. Where am I drawing a line from the origin to?
    Offline

    1
    ReputationRep:
    (Original post by thers)
    I don't understand. Where am I drawing a line from the origin to?
    Maybe an image of my own will help:

    Name:  image.jpg
Views: 88
Size:  80.3 KB

    Now we can see triangle A is the same as triangle B hence the angles theta and the one above which is unnamed in my image are equivalent, hence we can work out what theta is and double it.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Joshmeid)
    Maybe an image of my own will help:

    Name:  image.jpg
Views: 88
Size:  80.3 KB

    Now we can see triangle A is the same as triangle B hence the angles theta and the one above which is unnamed in my image are equivalent, hence we can work out what theta is and double it.
    Got it now. ty
 
 
 
Poll
Who is your favourite TV detective?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.