VCE Specialist 3/4 Question Thread!

[keltingmeith]

It's also worthwhile to note that Arg and arg are different. I'm not sure if you will lose marks if you left it in principle form when it asked for arg but just something to be aware of.

This happened at my school. During a SAC, I was the only one who noticed the distinction. I gained 3-6 marks on these questions over everybody else.

[Zues]

Got a few questions

• what does it mean by complex coefficients and real coefficients?
• what does factorise over C (complex field??) and over R mean?

and can anyone break this down a little please.
The fundamental theorem of algebra states that every polynomial P(z) with complex coefficients has at least one linear factor in the complex number system.
Any polynomial P(z) of degree n has n linear factors over C . Therefore a polynomial equation P ( z) = 0 of degree n has n solutions over C , provided that repeated solutions are counted theappropriate number of times.[/list][/list]

[Zues]

it is known that z = -3 + 2i is a solution to the equation z^3 + 5z^2 + 7z - 13 = 0. hence find all solutions to the equation

thanks

[Zues]

and

the solutions to the equation z^2 + mz + p = 0 are z = 3i and z = 2- 5i , find m and p.

also if you look at the attached , how does that happen?

thanks

[lzxnl]

So theres no need to understand the theory behind it bro?

Yeah you don't need this, but I'll say this for interest.
arg z is the imaginary part of the natural log of z
Now the inverse tan can be rewritten in terms of the natural log. Given that we want the inverse tan to spit out a certain set of numbers (-pi/2 to pi/2 exclusive), this puts a restriction on what we want the natural log to spit out and thus gives a natural restriction to arg z

Got a few questions

• what does it mean by complex coefficients and real coefficients?
• what does factorise over C (complex field??) and over R mean?

and can anyone break this down a little please.
The fundamental theorem of algebra states that every polynomial P(z) with complex coefficients has at least one linear factor in the complex number system.
Any polynomial P(z) of degree n has n linear factors over C . Therefore a polynomial equation P ( z) = 0 of degree n has n solutions over C , provided that repeated solutions are counted theappropriate number of times.[/list][/list]

So (1+i) x has a coefficient of x that is complex
Factorise over C means you can complex linear factorz like x+i
As for that theorem
it means any polynomial equation with one side 0 has a solution in the complex plane. Using factor theorem, you can divide your polynomial by a known linear factor to get a smaller degree polynomial. You can apply the theorem again and again to get more and more linear factors.

[keltingmeith]

it is known that z = -3 + 2i is a solution to the equation z^3 + 5z^2 + 7z - 13 = 0. hence find all solutions to the equation

thanks

Since what they've given is a solution, and the actual equation has real coefficients, we can then say that ANOTHER solution is the conjugate of the first solution - ie, z=-3-2i. From this, we can now say that , where a is some number. So, now all we need to do is find the value of a, using whatever method you like/deem necessary:



Equating coefficients, we get a=1. So, has solutions

and

the solutions to the equation z^2 + mz + p = 0 are z = 3i and z = 2- 5i , find m and p.

also if you look at the attached , how does that happen?

thanks

If we define f(z)=z^2+mz+p, then we know that f(3i)=0 and f(2-5i)=0. This means that we have a quadratic and two roots, which completely defines the function. So, let's take those roots and expand:


By equating coefficients, we get that m=2i-2 and p=6i+15

[Zues]

Since what they've given is a solution, and the actual equation has real coefficients, we can then say that ANOTHER solution is the conjugate of the first solution - ie, z=-3-2i. From this, we can now say that , where a is some number. So, now all we need to do is find the value of a, using whatever method you like/deem necessary:



Equating coefficients, we get a=1. So, has solutions
If we define f(z)=z^2+mz+p, then we know that f(3i)=0 and f(2-5i)=0. This means that we have a quadratic and two roots, which completely defines the function. So, let's take those roots and expand:


By equating coefficients, we get that m=2i-2 and p=6i+15

ok few questions, but before i get started, i see what you did with the conjugate question. What i sis was recognise that z = -3 + 2i was a solution and so was z = -3 - 2i, then i expanded these and got 9 + 4 = 13 then did (13) (z-a) and was like this has to be wrong

with step two and three of the first question, what did you do? could i have just expanded it and got the same answer, is this mathematically wrong?

with the second question, what i did was sub 3i and 2-5i into z and realised i need a third solution. is there another way of doing this question?

thanks

ps. would you also mind helping me what they did in the attachment above.

[keltingmeith]

ok few questions, but before i get started, i see what you did with the conjugate question. What i sis was recognise that z = -3 + 2i was a solution and so was z = -3 - 2i, then i expanded these and got 9 + 4 = 13 then did (13) (z-a) and was like this has to be wrong

I have no idea why you did this, or how it could've worked... You'd have to step me through your logic.

with step two and three of the first question, what did you do? could i have just expanded it and got the same answer, is this mathematically wrong?

You could've - I abused the fact that it was a difference of two squares to make my expanding life easier.

with the second question, what i did was sub 3i and 2-5i into z and realised i need a third solution. is there another way of doing this question?

You could've done that - I just tried it then and it worked. It took more time than what I did up there, though. You didn't need a third solution - the general rule is that to solve for n variables, you need n equations.

ps. would you also mind helping me what they did in the attachment above.

Sorry - I missed that amoung all your other questions.

[Zues]

4 becomes 2^2, and multiply everything inside the bracket by 2.

[Zues]

which should get you (2z + (4-2i)/(2) ) ^2, then you simplify and get (2z + 2-i ) ^2

[cosine]

which should get you (2z + (4-2i)/(2) ) ^2, then you simplify and get (2z + 2-i ) ^2

I know that, but the method im working with is completely legit and I mean, it should work... They both have the same amount of solutions, but my factored has the 2 outside the bracket. Does that make  a difference or nah?

[Conic]

This isn't a valid step. This leads to contradictions like when is actually equal to 32. If , then



You can't just say that because it isn't always true. In this example, you could say that and , so



The problem is that you are ignoring the squares when you are simplifying.

[Zues]

where have i gone wrong?

[Zues]

^^ i thought you can only complete the square with the coefficient of the x^2 is 1, they have done it leaving it with 4

[Conic]

After you get to you ignored the 4 when you used difference of two squares. You could have done something like



                               

                               

When you had you could have taken the 4 out as a factor.