Complex no's vector problems. (1 Viewer)

[Lumix]

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.

This is what I've done so far:

Z2 = Z1cis60 since it is an equilateral triangle

then I sub it in Z1^2 + Z2^2

I then get:

Z1^2 + (Z1cis60)^2

This is where I'm stuck...

I find vector problems very difficult and cannot solve most of them.. All the other topics in complex numbers I do fine except in vectors. Could someone give me some pointers on these questions?

 

[Lumix]

I'll add in another one...

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

 

[jayy100]

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

|z1| + |z2| >= |z1+z2| (its a fact- inequalities value)

We can manipulate that and write it as both:
i) |z1 + z2 - z1| =< |z1| + |z2-z1| ---> |z2| - |z1| =< |z2 - z1|
ii) | z2 + z1 - z2| =< |z2| + |z1-z2| --> |z1| - |z2| =< |21-z2|

therefore putting the latter 2 together you get:
||z2| - |z1|| =< |z1- z2|

i think thats right and sorry if it looks ugly! its really hard to help over the internet :S

 

[Lumix]

How come you are allowed to split the modulus if its addition? I thought only when it is |Z1Z2| then you can split to |Z1||Z2|

Edit:NVM LOL, I got this question now. Just need help on the first one!

 

[jayy100]

How come you are allowed to split the modulus if its addition? I thought only when it is |Z1Z2| then you can split to |Z1||Z2

umm i didnt. i just manipulated the inequalities value... (i think its possible LOL)

|z1 + z2 - z1| =< |z1| + |z2-z1|

you see this? the LHS essentially equals |z2|. So if you bring the |z1| on the RHS to the LHS you get the latter half... and you just do the same to ii)

but maybe someone else has a better method to do the question?

 

[jayy100]

and as for the first one, i know how to do it but its pretty long... and i dont like using latex or whatever it is....

but sketch a diagram to help you first...

If argz1 > arg z2,
then, z1= (alpha)z2 (where alpha is a complex number- since rotating is the same as multiplying by a complex number)
Note: then arg(alpha) = pie over 3 and modules of alpha = 1

then z1^2 + z2^2 = (az2)^2 + z2^2
= (alpha)^2(z2)^2 + z2^2
= z2^2 (alpha^2 + 1)

then a^2= cos(2pie/3) + isin (2pie/3)

and then you try and find an expression for alpha^2 + 1 and after some trig (which i can't type up properly, sorry)
you'll find that alpha^2 + 1 = 2 x 1/2 x alpha --> so alpha^2 + 1 = alpha

Hence, z1^2 + z2^2 = a x z2^2
= (alpha x z2) x z2 (i just re-arranged the order)
= z1 x z2 (Noting that we allowed z1 to equal (alpha x z2) in the beginning)

Similarly, if arg(z2) > arg (z1), inter-changing z1, z2, gives the same result.

Again, maybe there's another simpler method. Hope that helps tho :]

 

[Lumix]

and then you try and find an expression for alpha^2 + 1 and after some trig (which i can't type up properly, sorry)
you'll find that alpha^2 + 1 = 2 x 1/2 x alpha --> so alpha^2 + 1 = alpha

Can you tell me what trig identity u used to get 1+a^2 to become a?

 

[jayy100]

a^(2 )=cos 2pie/3 + isin 2pie/3
a^(2 )+ 1=(1 + cos 2pie/3)+ isin 2pie/3
=(cos^2 pie/3 + sin^2 pie/3 + cos^2 pie/3 - sin^2 pie/3)+ i×2sin pie/3 cos pie/3
=2cos pie/3 (cos pie/3+ isin pie/3)
=2 × 1/2 × a
=a
LOL, sorry for the delay, took me a while to type up.. AND LOL ask around for this question i guess? There should be a better way meee thinks :]

 

[hscishard]

This probably looks better
z1= rcis%
z2=rcis% x cis(pi/3)

z1^2 = r^2(cis%)^2
z2^2 = r^2(cis%)^2 x cis(pi/3)^2

z1^2 +z2^2 = r^2(cis%)^2 x [1+cis(2pi/3)]
=r^2(cis%)^2 x [1/2 + iroot3/2]
=r^2(cis%)^2 x cis(pi/3)

z1 x z2 = r^2(cis%)^2 x cis(pi/3)

 

[Lumix]

Thanks jay and hscishard! Cleared things up for me a lot!