VCE Specialist 3/4 Question Thread!

[Nobby]

Hey guys, some help for this thing would be great:

z² = a + bi
Let z = x + yi
Thus (x+yi)² = a + bi
x²-y² + 2xyi = a + bi
Therefore x²-y² = a
        and 2xy = b

Find x² in terms of a and b?

[pi]

x^2-y^2 = a   and    2xy = b
Hence, y = b/2x
=> y^2 = b^2/4x^2
Hence, x^2 - b^2/4x^2 = a
=> x^4 - b^2/4 = ax^2
=> x^4 - ax^2 - b^2/4 = 0

That's a quadratic for x^2, but I think the solution is going to be a bit ugly haha

[Nobby]

x^2-y^2 = a   and    2xy = b
Hence, y = b/2x
=> y^2 = b^2/4x^2
Hence, x^2 - b^2/4x^2 = a
=> x^4 - b^2/4 = ax^2
=> x^4 - ax^2 - b^2/4 = 0

That's a quadratic for x^2, but I think the solution is going to be a bit ugly haha

Yeah I got there but would there any neater ways to do it?
My working is going to look disgusting :S

EDIT: actually it looks okay

[Homer]

if is a factor of , how would i find the remaining factors?

[pi]

if is a factor of , how would i find the remaining factors?

Steps:
- Well, as (z-3i) is a factoryou then know that (z+3i) is also a factor.
- Hence, you would know that (z-3i)(z+3i)=(z^2 + 9) is a factor.
- From here, you can long divide or factorise in another way (your choice) to find the other quadratic factor
- Then factorise that (or jump to the quadratic formula) to find the final two linear factors

[Stick]

I'm not exactly sure, but I think you could apply long division, the factor theorem and the remainder theorem (much like cubics in Methods). Give it a go and let me know if it works.

EDIT: pi confirmed what I was thinking anyway. ^_^

[b^3]

Since the coefficients of the expression are real, then each complex factor will have a conjugate, that is also a factor, i.e. is also a factor. From that if you divide through by the two factors we have you should get a quadratic, that you can factorise to find the last 2 factors.

EDIT: Beaten, shouldn't have tried to spend time LaTeXing the long division before getting lazy...

[Planck's constant]

if is a factor of , how would i find the remaining factors?

The important piece of theory is that if a polynomial with REAL coefficient has a complex root, the conjugate of this root is also a root of the polynomial.

Here 3i is a root, therefore so is -3i
x-3i is a factor, x+3i is a factor
(x-3i)(x+3i) is a factor, therefore x^2 + 9 is a factor.

One way to proceed from here is long division of x^2 + 9 into the original polynomial.
The remainder is a quadratic which you can easily factorise

Alternatively, you can ignore my answer and go the one above

[Homer]

thanks heaps guys. I worked out the z^2+9 but i wasn't comfortable doing the long division. could someone please help me from there  

[BubbleWrapMan]

Let

Expanding gives


So
and

Hence

[Homer]

I get it. thanks guys

[Jaswinder]

Also by using the quadratic formula or by using the method of factorizing of polynomials in C,

(2z^2-4z+3) would break down into

so in total there are 4 solutions

[jadams]

Let

Expanding gives


So
and

Hence

Similarly, if you write out you can "by inspection" work out the other factor, by imagining what the expanded product of the two factors would look like. (Exact same method as that of Climbtoohigh, but could be used if you can manage the numbers in your head/scrap piece of paper, could possibly be quicker.)

[Homer]

If

show that

[Russ]

 Assuming , then , so the result holds iff