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Complex Number Q (1 Viewer)
- Thread starter OLDMAN
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underthesun
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hmm..
I'm not quite sure what maximize means in the sense of complex numbers..
I'm not quite sure what maximize means in the sense of complex numbers..
SoFTuaRiaL
Active Member
ok, but whats that got to do with the meaning of 'maximise'? is it the maximum value the the expression can have or something of that sort?
Must restate the problem.
If z is a unit complex number, ie. |z|=1, find the z that will give
|z^3-z+2| its greatest value.
This question is from the genre of questions described in Tip 19 of Geha : Maximum and Minimum values of |z|. As Geha has remarked, and I quote, "these questions test students' understanding of the geometry of the Argand diagram and are found to be more challenging.
A simple version might be: For |z-2|=1, find z when |z| takes its greatest value. It is helpful to draw an Argand diagram of the circle |z-2|=1 and graphically show that z=3 is the "complex number" that maximizes |z|.
However, be warned that the first problem stated is a bit more complicated than this.
If z is a unit complex number, ie. |z|=1, find the z that will give
|z^3-z+2| its greatest value.
This question is from the genre of questions described in Tip 19 of Geha : Maximum and Minimum values of |z|. As Geha has remarked, and I quote, "these questions test students' understanding of the geometry of the Argand diagram and are found to be more challenging.
A simple version might be: For |z-2|=1, find z when |z| takes its greatest value. It is helpful to draw an Argand diagram of the circle |z-2|=1 and graphically show that z=3 is the "complex number" that maximizes |z|.
However, be warned that the first problem stated is a bit more complicated than this.
N
ND
Guest
Can anyone tell me what 'cosa - cosb' equals? I've forget these trig identities and don't have a textbook with me.
N
ND
Guest
Anyway here's what i've got so far:
If |z| = 1, then z^3-z+2 = cis3@-cis@+2
= cos3@ - cos@ +2 + (sin3@ - sin@)i
.'. x = cos3@-cos@+2 = ....
y = sin3@-sin@ = ...
i imagine you could then find a locus, then differentiate to find max value, then work from there to find z. I'm not that sure though...
If |z| = 1, then z^3-z+2 = cis3@-cis@+2
= cos3@ - cos@ +2 + (sin3@ - sin@)i
.'. x = cos3@-cos@+2 = ....
y = sin3@-sin@ = ...
i imagine you could then find a locus, then differentiate to find max value, then work from there to find z. I'm not that sure though...
N
ND
Guest
Why is it squared?
I was wondering why this question is finding it difficult to liftoff. Perhaps it is my notation |z| which refers to mod(z). Since
|cos3@ - cos@ +2 + (sin3@ - sin@)i|=sqrt((cos3@ - cos@ +2)^2 + (sin3@ - sin@)^2) it might just be simpler to work with |....|^2.
The real challenge to this question is how to efficiently find the maxima, this part I daresay makes this question a "level 8" ; my original suggestion differentiate with respect to @, may prove to messy. I have found a much neater way , get everything in terms of cos@, and treat the resulting cubic polynomial in cos@ as a simple cubic polynomia in c, where c=cos@. Find the turning points of this cubic polynomial and show one of them is the maximum. Woudn't mind bringing in Spice Girl into the discussion, when ND's given it a further go.
|cos3@ - cos@ +2 + (sin3@ - sin@)i|=sqrt((cos3@ - cos@ +2)^2 + (sin3@ - sin@)^2) it might just be simpler to work with |....|^2.
The real challenge to this question is how to efficiently find the maxima, this part I daresay makes this question a "level 8" ; my original suggestion differentiate with respect to @, may prove to messy. I have found a much neater way , get everything in terms of cos@, and treat the resulting cubic polynomial in cos@ as a simple cubic polynomia in c, where c=cos@. Find the turning points of this cubic polynomial and show one of them is the maximum. Woudn't mind bringing in Spice Girl into the discussion, when ND's given it a further go.
underthesun
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not sure if you still need it, but
cos(A) - cos(B) = -2sin((A+B)/2)sin((A-B)/2)
I managed to memorise these, but I'm not sure if they'll be of any help during the any tests?
turtle_2468
Member
It's bashy, but it comes out eventually...
oldman: if you differentiate with respect to @ it's still not too bad, you just need to use a trig identity (I think you're trying not to give the soln, hence I'm not being too specific here) to factor out an obvious factor, then the turning pts (or rather the cos @ values that create them) come out quite easily.
oldman: if you differentiate with respect to @ it's still not too bad, you just need to use a trig identity (I think you're trying not to give the soln, hence I'm not being too specific here) to factor out an obvious factor, then the turning pts (or rather the cos @ values that create them) come out quite easily.
turtle :
True it really comes out the same if you convert all trig. in terms of cos@. Thus you might have f(C(@))=|...|^2 where C(@)=cos(@)
Thus if you differentiate with respect to @, you can get
df(C(@))/d@=(df(C)/dC)*dcos@/d@=-(df(C)/dC)*sin@ (chain rule!)
so we are really back to the polynomial in C, with additional stationary points 0 and pi, and whatever the stationary pts the polynomial yields.
True it really comes out the same if you convert all trig. in terms of cos@. Thus you might have f(C(@))=|...|^2 where C(@)=cos(@)
Thus if you differentiate with respect to @, you can get
df(C(@))/d@=(df(C)/dC)*dcos@/d@=-(df(C)/dC)*sin@ (chain rule!)
so we are really back to the polynomial in C, with additional stationary points 0 and pi, and whatever the stationary pts the polynomial yields.
turtle_2468
Member
lol yeah! The chain rule... now why didn't I think of that 
must be getting rusty. BTW the polynomial factorises quite nicely
must be getting rusty. BTW the polynomial factorises quite nicelyspice girl
magic mirror
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the question should be why are some of the mathematicians duped into doing med...
turtle_2468
Member
lol! just spice girl and I I think some ppl have trouble solving x/15+x=constant... :-( oh well, that's life I guess.
some ppl have trouble solving x/15+x=constant... :-( oh well, that's life I guess.
N
ND
Guest
Well i had a go using oldman's suggestion, but when i expanded f(a)=|...|^2 where a=cos@ i got:
f(a) = 16a^3 - 3a^2 - 16a + 7
f'(a) = 48a^2 - 6a - 16, which doesn't have any rational solutions for f'(a)=0.
f(a) = 16a^3 - 3a^2 - 16a + 7
f'(a) = 48a^2 - 6a - 16, which doesn't have any rational solutions for f'(a)=0.
N
ND
Guest
Hehe, i just hope i can pass section III of the UMAT. Damn shapes...