Further Maths AS-level complex numbers question

I'm a bit confused on how to approach this question. I was thinking of substituting one of the root into the function to find the value of Q and then using long division to see if it's a factor of f(z), but that would take sooo long for only 4 marks. Any guidance would be appreciated 😊.
f(z) = z^4 -10z^3 +71z^2+Qz +442, where Q is a real constant.
Given that z=2-3i is a root of the equation f(z)=0
a) show that z^2 -6z +34 is a factor of f(z) (4 marks)

(edited 8 months ago)

Reply 1

JG0005

8

Original post by Enkindle

I'm a bit confused on how to approach this question. I was thinking of substituting one of the root into the function to find the value of Q and then using long division to see if it's a factor of f(z), but that would take sooo long for only 4 marks. Any guidance would be appreciated 😊.
f(z) = z^4 -10z^3 +71z^2+Qz +442, where Q is a real constant.
Given that z=2-3i is a root of the equation f(z)=0
a) show that z^2 -6z +34 is a factor of f(z) (4 marks)

i havent done maths for a few months now so im a bit rusty but i think you just do long division, then if the final line is =0 then its a factor? at least thats how i was taught it. hope this helps!

Original post by Enkindle

I'm a bit confused on how to approach this question. I was thinking of substituting one of the root into the function to find the value of Q and then using long division to see if it's a factor of f(z), but that would take sooo long for only 4 marks. Any guidance would be appreciated 😊.
f(z) = z^4 -10z^3 +71z^2+Qz +442, where Q is a real constant.
Given that z=2-3i is a root of the equation f(z)=0
a) show that z^2 -6z +34 is a factor of f(z) (4 marks)

That's an approach but as you suspect it's not the optimal one.

The most optimal approach is using the fact that every coefficient of

f

is real which must mean roots come in conjugate pairs. So if

z=2-3i

is a root you can immediately state another one.

So if

z_1,z_2

are roots of

f

then

(z-z_1)(z-z_2)

is a factor of

f

.

This is about four marks. The only computational bit of the solution is multiplying out a pair of brackets.

EDIT: I reread the question and I realise that what I had just put down is only half the story. Multiplying the two factors will give a quadratic but it’s not the one you are after, so you would technically (possibly) divide the quartic by the quadratic you have just found to get the quadratic you are being asked for. This is complicated by not knowing Q, however.

The slick approach instead is to note that your quartic has four roots

z_1,z_2,z_3,z_4

.

By considering sum and products of roots on the quartic, it should hopefully be clear that:

z_1+z_2+z_3+z_4 = 10

…. (*)

z_1z_2z_3z_4 = 442

…. (**)

and we can form similar relations concerning coefficients

71,Q

however we do not actually need these.

From what I said we know two of the roots

z_1,z_2

which means our quadratic has roots

z_3,z_4

so that it takes on the form

z^2+Az+B

where

z_3+z_4 = -A

z_3z_4 = B

So if we can determine what the sum

z_3+z_4

is and also what the product

z_3z_4

then we know what

A,B

are hence we know the quadratic.

Using (*) we find that

z_3+z_4 = 10 - z_1 - z_2

and this is our

-A

Using (**) we further find that

z_3z_4 = \dfrac{442}{z_1 z_2}

and this is our

B

.

The neat thing is that we never even need to know what Q is with this method, nor what the actual other two roots are individually.

(edited 8 months ago)

Reply 3

JG0005

8

Original post by RDKGames

That's an approach but as you suspect it's not the optimal one.

The most optimal approach is using the fact that every coefficient of

f

is real which must mean roots come in conjugate pairs. So if

z=2-3i

is a root you can immediately state another one.

So if

z_1,z_2

are roots of

f

then

(z-z_1)(z-z_2)

is a factor of

f

.

This is four marks. The only computational bit of the solution is multiplying out a pair of brackets.

yeah youre right, thats another way to do it but tbh you just do whats best for you. it doesnt really matter (unless youre running out of time in an exam then obviously use the quicker way), as long as it gets you to the final answer, then perfect. i used my method in my finals and got an A, you seem to have a good understanding about what youre doing though so either way is fine.

Original post by JG0005

yeah youre right, thats another way to do it but tbh you just do whats best for you. it doesnt really matter (unless youre running out of time in an exam then obviously use the quicker way), as long as it gets you to the final answer, then perfect. i used my method in my finals and got an A, you seem to have a good understanding about what youre doing though so either way is fine.

Your method only works well if we know Q already. Then yes, without even knowing a single root we can just do long division to get 0 remainder.

Unfortunately we have the added difficulty of not knowing Q, and trying to determine it can turn into a very lengthy process. It’s good to be aware of the short solutions and train what to look out for in similar contexts.

(edited 8 months ago)

Reply 5

JG0005

8

Original post by RDKGames

Your method only works well if we know Q already. Then yes, without even knowing a single root we can just do long division to get 0 remainder.

Unfortunately we have the added difficulty of not knowing Q, and trying to determine it can turn into a very lengthy process. It’s good to be aware of the short solutions and train what to look out for in similar contexts.

oh sorry sorry, i forgot to add that part. if i remember correctly, i was taught to do long division then create a quadratic equation from the answer to find q. this type of question was one of the favourites because it was such an easy 4 marks. as i said, its been a while since i took maths so if my method is wrong please correct me. i hope i havent confused you Enkindle

Further Maths AS-level complex numbers question

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