You are Here: Home >< Maths

# Further Maths Complex Numbers watch

1. I've been given the question:
One root of the quadratic equation z^2 + az + b = 0 where a and b are real, is the complex number 1 + 2i.
i) Write down the other root
ii) Find the values of a and b
I'm unsure how to answer this one.

I also got the question:
Find the quadratic equation which has roots 2 + 3i and 2 - 3i.
For this I expanded (2+3i)(2-3i) and ended up with 13, which is obviously wrong, but I can't think of another way of doing it.

Hopefully someone can help.
2. (Original post by CCauston113)
I've been given the question:
One root of the quadratic equation z^2 + az + b = 0 where a and b are real, is the complex number 1 + 2i.
i) Write down the other root
ii) Find the values of a and b
I'm unsure how to answer this one.

I also got the question:
Find the quadratic equation which has roots 2 + 3i and 2 - 3i.
For this I expanded (2+3i)(2-3i) and ended up with 13, which is obviously wrong, but I can't think of another way of doing it.

Hopefully someone can help.
First question: i) if 1+2i is a root, then 1-2i is also a root as complex numbers always come in conjugate pairs because of the quadratic formula ii) both 1+2i and 1-2i are a root, so z=1±2i. To find the equation: (z-(1+2i))(z-(1-2i)) = (z-1-2i)(z-1+2i) then expand the brackets, all the i will cancel out and you will be left with z^2 - 2z + 4 = 0, so compare this to z^2 + az + b = 0, and a = -2 and b = 4 Second question: do the same as above, so we know the roots are z=2±3i, so make brackets: (z-(2+3i))(z-(2-3i)) = (z-2-3i)(z-2+3i) and expand out so all the i cancel and you are left with a quadratic - which I have worked out but tell me what you get first and we can compare hope this helps
3. (Original post by megfield22)
First question: i) if 1+2i is a root, then 1-2i is also a root as complex numbers always come in conjugate pairs because of the quadratic formula ii) both 1+2i and 1-2i are a root, so z=1±2i. To find the equation: (z-(1+2i))(z-(1-2i)) = (z-1-2i)(z-1+2i) then expand the brackets, all the i will cancel out and you will be left with z^2 - 2z + 4 = 0, so compare this to z^2 + az + b = 0, and a = -2 and b = 4 Second question: do the same as above, so we know the roots are z=2±3i, so make brackets: (z-(2+3i))(z-(2-3i)) = (z-2-3i)(z-2+3i) and expand out so all the i cancel and you are left with a quadratic - which I have worked out but tell me what you get first and we can compare hope this helps
OK, thanks
4. (Original post by CCauston113)
OK, thanks
For the first one shouldn't it be z^2 - 2z + 5?

For the second one I got z^2 - 4z + 13.
5. (Original post by CCauston113)
For the first one shouldn't it be z^2 - 2z + 5?

For the second one I got z^2 - 4z + 13.
yes it is, thanks for spotting it was just a typo I think because I have +5 written down in my notes! thanks

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 12, 2017
Today on TSR

### Edexcel GCSE Maths Unofficial Markscheme

Find out how you've done here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams