VCE Specialist 3/4 Question Thread!

[brightsky]

I tend not to use the discriminant. All you really need to do is complete the square and see whether the constant at the end is positive or negative. The examples below should illustrate what I mean.

Example with a Positive Constant: z^2 + z + 1 = (z^2 + z + 1/4) - 1/4 + 1 = (z + 1/2)^2 + 3/4 = (z + 1/2)^2 - 3/4 i^2 = (z + 1/2)^2 - [sqrt(3)/2 i]^2 = (z + 1/2 - sqrt(3)/2 i) (z + 1/2 + sqrt(3)/2 i)

Example with a Negative Constant: z^2 - 2z - 1 = (z^2 - 2z + 1) - 1 - 1 = (z - 1)^2 - 2 = (z - 1)^2 - [sqrt(2)]^2 = (z - 1 - sqrt(2))(z - 1 + sqrt(2))

As you can see, after completing the square, if you get a positive constant at the end, then there will be two complex linear factors, and if you get a negative constant, then there will be two real linear factors.

[keltingmeith]

What do you mean factorise if by sight? Just remember what you see is different to what i see in an expression haha

For example:



By sight, I can see that this has linear factors (x-4)(x-1) (because they are multiples of 4 and add to -5), so:



Now, let's consider:



Now, I can't figure out any linear factors just by looking at this - so instead, I use the quadratic equation:



And so, I see that this quadratic only has complex solutions.

[cosine]

For example:



By sight, I can see that this has linear factors (x-4)(x-1) (because they are multiples of 4 and add to -5), so:



Now, let's consider:



Now, I can't figure out any linear factors just by looking at this - so instead, I use the quadratic equation:



And so, I see that this quadratic only has complex solutions.

Thank you man, cleared that up for me heaps! I mean, i can do them dont get me wrong but its just that I couldnt figure why some expressions did not need to be introduced to 'i'.

But once again, thank you. Really appreciate the help!

[Zues]

how do i do "b" and "c". again with b i dont know how to long divide with brackets involved. i know you long divide with the factor (z-1), but how?

as for c, i have no idea.

thanks

[brightsky]

 b) Excuse the formatting, but:

              z^3  -    3i z^2   -    3z          + (4+i)
            ----------------------------------------------------------
 z - 1   |  z^4 - (1+3i)z^3 + 3(i-1)z^2 + (7+i)z - 4 - i
              z^4 -  z^3
            ----------------------
                        - 3i z^3    + 3(i-1)z^2
                        - 3i z^3    + 3i z^2
                       -----------------------------
                                            -3z^2     + (7+i)z
                                            -3z^2     +   3z
                                            ----------------------
                                                              (4+i)z - 4 - i
                                                              (4+i)z - 4 - i
                                                             -----------------
                                                                               0

Hence, Q(z) = z^3 - 3i z^2 - 3z + (4+i).

c) By inspection, Q(z) = (z - i)^3 + 4. Completing the cubic is, in general, quite difficult, as is evident from the complexity of cubic formula: http://www.math.vanderbilt.edu/~schectex/courses/cubic/. But in this case, we can see that z^3 - 3iz^2 - 3z looks somewhat like the expansion of (z-i)^3. All we need to do adjust (z-i)^3 by adding 4 to obtain Q(z).

[Zues]

b) Excuse the formatting, but:

              z^3  -    3i z^2   -    3z          + (4+i)
            ----------------------------------------------------------
 z - 1   |  z^4 - (1+3i)z^3 + 3(i-1)z^2 + (7+i)z - 4 - i
              z^4 -  z^3
            ----------------------
                        - 3i z^3    + 3(i-1)z^2
                        - 3i z^3    + 3i z^2
                       -----------------------------
                                            -3z^2     + (7+i)z
                                            -3z^2     +   3z
                                            ----------------------
                                                              (4+i)z - 4 - i
                                                              (4+i)z - 4 - i
                                                             -----------------
                                                                               0

Hence, Q(z) = z^3 - 3i z^2 - 3z + (4+i).

c) By inspection, Q(z) = (z - i)^3 + 4. Completing the cubic is, in general, quite difficult, as is evident from the complexity of cubic formula: http://www.math.vanderbilt.edu/~schectex/courses/cubic/. But in this case, we can see that z^3 - 3iz^2 - 3z looks somewhat like the expansion of (z-i)^3. All we need to do adjust (z-i)^3 by adding 4 to obtain Q(z).

thanks brightsky, however im still not sure how to long divide with brackets? i.e. what do i minus? what do i look to see what my next part of the quotient is for example in - (1+3i)z^3 ? and also when i minus z^3 what do i minus it from since i still have that 1 there?

thanks

[Zues]

how to do this?

[cosine]

how to do this?

Complex roots always come in conjugate pairs.
So if and  
But if they're in polar form, z =   then the conjugate of z is

So your answer is C

[Zues]

it also talks about using pi + pi/8 because the roots have equal spacing. i understand why pi is added, e.g. splitting in half, but why is it adde to pi/8 as apposed to 7pi/8?

[keltingmeith]

Complex roots always come in conjugate pairs.
So if and  
But if they're in polar form, z =   then the conjugate of z is

So your answer is C

Not true. This only works for factorising, where all coefficients are real.

it also talks about using pi + pi/8 because the roots have equal spacing. i understand why pi is added, e.g. splitting in half, but why is it adde to pi/8 as apposed to 7pi/8?

I'm assuming because they messed up - it should be 7pi/8, i.e B.

[Zues]

Not true. This only works for factorising, where all coefficients are real.

I'm assuming because they messed up - it should be 7pi/8, i.e B.

how does this work? can you show me with steps

[cosine]

Not true. This only works for factorising, where all coefficients are real.

Can you elaborate?

[brightsky]

Can you elaborate?

The Conjugate Root Theorem states that if P(x) is a polynomial in x with all real coefficients, then the roots of P(x) occur in conjugate pairs; that is, if a + bi is a root of P(x), where a, b E R, then a - bi is also a root of P(x).

Note that the Conjugate Root Theorem only applies to polynomials with all real coefficients. If there is even one coefficient that is not real, the Conjugate Root Theorem does not work.

[cosine]

The Conjugate Root Theorem states that if P(x) is a polynomial in x with all real coefficients, then the roots of P(x) occur in conjugate pairs; that is, if a + bi is a root of P(x), where a, b E R, then a - bi is also a root of P(x).

Note that the Conjugate Root Theorem only applies to polynomials with all real coefficients. If there is even one coefficient that is not real, the Conjugate Root Theorem does not work.

\

Sorry to be nooby, but how do we know that this specific questions consists of un real coefficients of z? Im confused now :/

[keltingmeith]

\

Sorry to be nooby, but how do we know that this specific questions consists of un real coefficients of z? Im confused now :/

So, the trick is that this isn't a polynomial - it's just a complex number with unknown modulus. So, we cannot apply the conjugate root theorem, because that only applies to polynomials.