two quick maths questions for the pros

[headRush]

well to be honest, im completely stuck on these. help is greatly appreciated, a worked solution would be exquisite.

find all complex numbers z, such that:

az^2 + b(z bar) + c = 0

where z bar is the conjugate of z, and b^2 - 4ac > 0

and.....!

find all complex numbers z such that Re(z^3) < 0, show the solutions graphically in the complex plane. hint: use polar coordinates

 

[JJChicken]

let z=A+Bi where A,B are real constants and i is well iota

thus az^2+b(z bar) + c = 0 becomes

a(A^2 + 2ABi - B^2) + b(A - Bi) + C = 0 since i^2 =-1 and I think (now its been four years since I did this so I'm likely long) that z bar = A - Bi

Since =0, therefore terms with "i" must equal 0 and terms without "i" must equal 0 as well (0=0+0i), jusst solve for that ya

As for the second one, use the same approach - z = A + Bi

Therefore, z^3 = (A^2 +2ABi - B^2)(A+Bi) = A^3 +2(A^2)Bi - (B^2)A +(A^2)Bi-2(B^2)A - (B^3)i

Collate terms without "i", and the solution is that <0, i.e. A^3 - 3(B^2)A < 0

The graph will be a cubic (edited: before I WRONGLY said it would be a parabola) methinks.

I might have made some silly mistakes in my workings but the method is 100% correct as long as z bar = A - Bi which im prety sure it is.

Cheers :beer:

 

[headRush]

well i've been playing with it for the past few hours (yes, bad mathematician 🙁)

i used the same basic approach as you, except it isn't all equal to 0,

i let z = x+iy (instead of a+ib because we already have a's and b's and its all very confusing 😛)

i said

a(x+iy)^2 + b(x-iy) + c = 0

we then expand it

a(x^2 - y^2 + 2xyi) + bx - byi + c = 0

and equate the real and imaginary parts to the left hand side, which is zero, setting up two simultaneous equations, ie:

a(x^2 - y^2) + bx + c = 0
a(2xy) - by = 0

then we can see that

y(2ax-b) = 0

the trivial case in which y = 0 turns the original equation into

a(x+0i)^2 + b(x-0i) + c = 0

a regular quadratic in x^2, so i discarded this solution

the alternate solution is that:

2ax - b = 0

thus x = b/2a

then i substitute b/2a for x into a(x^2 - y^2) + bx + c = 0

but i cannot find a clean and discrete solution for y 🙁

i have made a genuine attempt at the problem but am not really getting anywhere, so can any of you atot maths geniuses help me out here?

 

[olds]

Huh, I didn't think "maths" was a word but it appears to be one.

 

[DrPizza]

I didn't think too much about it, but think about the hint for polar coordinates. Maybe a solution is easier to work out using the trig form of the complex numbers, rather than using rectangular coordinates.

i.e. r = sqrt(a²+b²) and theta = arctan(b/a)
It's very easy to multiply, square, and take roots of complex numbers in the form r(cos(theta) + i sin(theta))

But, I can't recall doing a similar problem & don't really have the time at the moment to work it out to see if this approach would be easier/quicker.

 

[headRush]

well i've solved the first part and it worked out fine, there is no, nice or elegant looking, discrete solution. although i invite others to try.

the part for re(z^3) < 0 is still open, i cannot do it at all.

 

[JJChicken]

Originally posted by: headRush
well i've solved the first part and it worked out fine, there is no, nice or elegant looking, discrete solution. although i invite others to try.

the part for re(z^3) < 0 is still open, i cannot do it at all.

Remember for your first part, where you have y(2ax-b) = 0 , you have two possibilities:

(i) y = 0
(ii) x = b/2a

Both are EQUALLY GOOD, there's no need (and it would be incorrect) to "discard" the regular quadratic in x^2 for (i). Basically your answer will look like:

If y=0, ....

If x=b/2a, ...

You mention that you solved it, which is great news, just keep in mind not to discard the y=0 solution because its a quadratic. Also, do you remember how to solve quadratics? You should get two values of x from it.

Now onto the Re(z^3) bit, can you provide your workings? I think the answer is in my above post, but it may be better for me to explain it differently. So if you can provide what you are doing, then I can try to find your stumbling block and explain it differently.

Cheers :beer:

 

[dawp]

42

 

[JJChicken]

Originally posted by: DrPizza
I didn't think too much about it, but think about the hint for polar coordinates. Maybe a solution is easier to work out using the trig form of the complex numbers, rather than using rectangular coordinates.

i.e. r = sqrt(a²+b²) and theta = arctan(b/a)
It's very easy to multiply, square, and take roots of complex numbers in the form r(cos(theta) + i sin(theta))

But, I can't recall doing a similar problem & don't really have the time at the moment to work it out to see if this approach would be easier/quicker.

This is a good idea as well.

Ahh brings back memories