complex numbers question (1 Viewer)

[Rahul]

here is a q i have been having troubles wid:

on a argand diagram the points P and Q represent the number z1 and z2. OPQ is an equilateral triangle. Show that z1^2 + z2^2 = z1.z2

any hints or suggestions?

btw it is from the cambridge 4u book, so the guys wid the worked solutions may be able to help out here. it is Ex 2.3, q5.

 

[McLake]

Here's my sol:

|z1| = |z2| = r

arg(z1) = arg(z2) + pi/3
let arg(z1) = @
so arg(z2) = @ + pi/3

z1^2 + z2^2 = (rcis@)^2 + (rcis(@+pi/3))^2
z1^2 + z2^2 = r^2cis2@ + r^2cis2(@+pi/3)
z1^2 + z2^2 = r^2[cos2@ + isin2@] + r^2[cos(2@+2pi/3) + isin(2@+2pi/3)]
z1^2 + z2^2 = r^2[cos2@ + cos(2@+2pi/3) + isin(2@+2pi/3) + isin2@]
z1^2 + z2^2 = r^2[cos2@ + cos2(@+pi/3) + isin2(@+pi/3) + isin2@]

Not sure if this is allowed, but it seems logical

z1^2 + z2^2 = [cos2(2@+pi/3) + isin2(2@+pi/3)]

Now:
z1*z2 = (rcis@)(rcis(@+pi/3))
z1*z2 = r^2[(cis@)(cis@+pi/3)]
z1*z2 = r^2[(cos@ + isin@)(cos{@+pi/3} + isin{@+pi/3}]
z1*z2 = r^2[(cos@*cos{@+pi/3} + cos@*isin{@+pi/3} + isin@*cos{@+pi/3} - sin@*sin{@+pi/3}]
z1*z2 = r^2[cos2(@ + (@ + pi/3))
+ sin2(@ + (@ + pi/3))]
z1*z2 = r^2[cos2(2@ + pi/3)
+ isin2(2@ + pi/3)]

Now they are both equal ....


NB: If that step ISN'T allowed then just expand the cos2 and sin2 terms. If any one wants me to do this then just let me know ...

 

[OLDMAN]

An heroic effort McLake.

Another solution:

Z2=Z1cis(pi/3)
Z2^2=Z1^2cis(2pi/3)
Z1^2+Z2^2=Z1^2(1+cis(2pi/3))

Now 1+cis(2pi/3)=cis(pi/3)

Therefore Z1^2+Z2^2=Z1Z2 proven.

 

[McLake]

Originally posted by OLDMAN
An heroic effort McLake.

Another solution:

Z2=Z1cis(pi/3)
Z2^2=Z1^2cis(2pi/3)
Z1^2+Z2^2=Z1^2(1+cis(2pi/3))

Now 1+cis(2pi/3)=cis(pi/3)

Therefore Z1^2+Z2^2=Z1Z2 proven.

DAMN, I always seem to overcomplicate things ...

 

[OLDMAN]

An important Trigonometry note arises out of McLake's sol.

All trigonometry identities are learnt in 3Unit, True or False.

False. There is an important set that's quite unigue to 4U:
sina+sinb=2sin((a+b)/2)cos((a-b)/2)
sina-sinb=2sin((a-b)/2)cos((a+b)/2)
cosa+cosb=2cos((a+b)/2)cos((a-b)/2)
cosa+cosb=-2sin((a+b)/2)sin((a-b)/2)

The next question is :How does the examiner expect me to memorize all these?

You don't. Work them out from scratch. Two numbers 2 and 8 could be written in "midpoint+or-displacement" 5-2 and 5+3. Thus two numbers a,b could be (a+b)/2+(a-b)/2,(a+b)/2-(a-b)/2

Watch: cos(2pi/3)+cos(0) midpoint of 2pi/3 and 0 is pi/3, magnitude of displacement is pi/3.
Thus, cos(pi/3+pi/3)+cos(pi/3-pi/3)=
cos(pi/3)cos(pi/3)-sin(pi/3)sin(pi/3)+cos(pi/3)cos(pi/3)+sin(pi/3)sin(pi/3)=2cos(pi/3)cos(pi/3) which is the formula.

Another example sin8+sin2=sin(5+3)+sin(5-3)
=sin5cos3+cos5sin3+sin5cos3-cos5sin3=2sin5cos3

There are always 2 terms in the expansion that cancel each other out.