Geometrical representation of a complex number as a vector (1 Viewer)

[YBK]

Hey, just a few questions from Cambridge 2.3; any help would be greatly appreciated.

5. On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z1^2 + z2^2 = z1z2

(could you please elaborate on the first step, because I have the answer to it, I'm just not sure how the teacher got it)


6. Show that | |z1| - |z2| | <or equal to |z1 + z2|
State the condition for equality to hold.


8. Show that | z1 + z2 + ... + zn | <or equal to |z1| + |z2| + ... + |zn|




Thanks

 

[ianc]

Hey, just a few questions from Cambridge 2.3; any help would be greatly appreciated.

5. On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z1^2 + z2^2 = z1z2

(could you please elaborate on the first step, because I have the answer to it, I'm just not sure how the teacher got it)


<or equal="" to="" |z1="" +="" z2=""><or equal="" to="" |z1="" |z2="" .="" +="" |zn="">

A couple of points:
</or></or>

• <or equal="" to="" |z1="" +="" z2=""><or equal="" to="" |z1="" |z2="" .="" +="" |zn=""> O is the origin of the argand diagram</or></or>
• OP = OQ (equilateral triangle), and hence modulus of z1 = modulus of z2
• The angle between OP and OQ is π/3 (equilateral triangle again)

You could probably prove this question algebraically, but for me that would take a long time as you would have to expand z1 and z2 out using de moivre's theorem. To start on that method you'd let z1 = rcis(θ<o></o>) and z2 = rcis(θ<o></o> + π/3)

Hence, I would solve it graphically (I prefer doing this anyway). Here's a few tips on how you'd actually draw it:

• Draw point Q on the x axis (this makes it a little simpler)
• Draw point P, making sure it is π/3 away from Q and their moduli are equal
• Draw the point Q^2, which should be easy since its argument is 0 and you only need to square the modulus
• Draw P^2, remembering to add π/3 to the argument
• Then draw the point P^2 + Q^2 (remember you use a parallelogram to add complex nos)
• Then draw the point P*Q, which will be the same point as P^2+Q^2

6. Show that | |z1| - |z2| |
State the condition for equality to hold.

8. Show that | z1 + z2 + ... + zn |

I don't think you typed these questions out fully.



Once you get used to the geometrical complex number questions, they become easy.

Hope this helps.

 

[YBK]

Thank you, ianc!!




6. Show that | |z1| - |z2| | < or equal to |z1 + z2|

State the condition for equality to hold.

8. Show that | z1 + z2 + ... + zn | < or equal to |z1| + |z2| + ... + |zn|




Though I'm 100% sure that I completely typed out the question. Weird that it didn't show up.

 

[ianc]

No worries about the typo, when I was typing the previous post I clicked preview first and it had somehow replaced the theta symbol with emoticons.

Anyway...

6. Show that | |z1| - |z2| | < or equal to |z1 + z2|

State the condition for equality to hold.

• Draw any two points on an argand diagram, z1 and z2. Call them A and B respectively
• Draw z1+z2, call it point C.
• OB = AC (parm)
• Working in triangle ACO, we know that CO is always greater than AC or AO.
• Hence AO - AC (or |z1| - |z2|) will always be less than CO (which is |z1 + z2|
• When it says what condition is needed for the equality to hold, that means when does the lefthandside = righthandside
• In this case, it is when z1 = -z2, or vice versa (i think!)

8. Show that | z1 + z2 + ... + zn | < or equal to |z1| + |z2| + ... + |zn|

How I would do this one:

• Draw a few points on the argand diagram
• Measure all the moduli, and add them up simply by a straight line. This point represents |z1| + |z2| + .... + |zn|
• Add up all the complex numbers you drew (using several parms) This point represents | z1 + z2 + ... + zn |
• Use a compass to rotate the point representing the total across to the straight line.
• The compass will rest on a point on the straight line, so hence the modulus is smaller, so you have proved it.

Hopefully you understood the reasoning behind this, because it's probably a pretty dicey method (using the compass and having to measure )


Anyway, glad to be of assistance