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Roots of Equations Further Maths Help

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I need to use the fact that it has exactly one root to find the y part - do I use the quadratic formula on the last bit I have written?

I have x already and z = x+iy

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Reply 1

Original post
by wbanner2001
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I need to use the fact that it has exactly one root to find the y part - do I use the quadratic formula on the last bit I have written?

I have x already and z = x+iy

Could you make The question a bit larger/clearer pls.

Reply 2

Original post
by mqb2766
Could you make The question a bit larger/clearer pls.


zz* + 3iz = p + 9i
where z is a complex number and p is a real constant.
Given that this equation has exactly one root, determine the complex number z

Reply 3

Original post
by wbanner2001
zz* + 3iz = p + 9i
where z is a complex number and p is a real constant.
Given that this equation has exactly one root, determine the complex number z

Not checked your working but if there is only one solution, your quadratic in y must have a single soln, so discriminant is zero or complete the square (same thing ...)

Reply 4

Original post
by mqb2766
Not checked your working but if there is only one solution, your quadratic in y must have a single soln, so discriminant is zero or complete the square (same thing ...)

Thanks
I got my final answer as z = 3 + 3/2 i

Reply 5

Original post
by wbanner2001
Thanks
I got my final answer as z = 3 + 3/2 i

Presume you've checked by subbing it back into the equation?

Reply 6

Original post
by mqb2766
Presume you've checked by subbing it back into the equation?


Yeah - all good:smile:

Reply 7

Original post
by mqb2766
Presume you've checked by subbing it back into the equation?


Could you check where I have gone wrong on this proof by induction stage 3:

B9D12B52-67C2-4BCF-9FEB-E3BFC1F3C1AC.jpg.jpeg6C8FC084-9463-44A4-9419-9671E3F057E5.jpg.jpeg
A4FC9619-FC25-4197-B869-F1E4D971AEF6.jpg.jpegD5F56943-A0E3-4F39-B1C4-361D8C192630.jpg.jpeg

Reply 8

When you add the k+1^th term, you multiply by 3/3 to get a 1/3 outside the factors. However only one of the factors should be multiplied by 3. If you multiply them both you're doing 9/3.

Reply 9

Original post
by mqb2766
When you add the k+1^th term, you multiply by 3/3 to get a 1/3 outside the factors. However only one of the factors should be multiplied by 3. If you multiply them both you're doing 9/3.


Ah fab! Done it now :smile:

Reply 10

Sorry to bother you again but am I doing the right thing here needing to find the tangent:


C854975D-5D20-4389-B079-49439B443A89.jpg.jpeg

Reply 11

What are you doing, also what is the first part of the question?

Reply 12

Original post
by mqb2766
What are you doing, also what is the first part of the question?


Completed that but I can’t work out why my values for a here are inconsistent:

2870DEA0-C4DA-4F49-A5A4-EAF1D64BFFFD.jpg.jpeg

Reply 13

Write both equations as
a = ...
Then equate and solve.

Reply 14

I did that originally - I just made an arithmetic error putting k=2 not 1/2 aha! Thanks for your help!

Reply 15

I'm a year late but do you know what paper this is from?

Reply 16

Original post
by rrrjjfg
I'm a year late but do you know what paper this is from?

Thats what i want to know

Reply 17

Original post
by wbanner2001
Attachment not found


I need to use the fact that it has exactly one root to find the y part - do I use the quadratic formula on the last bit I have written?

I have x already and z = x+iy

If you still remember what paper this is, please let us know!

Reply 18

Original post
by rrrjjfg
If you still remember what paper this is, please let us know!

Hi, I left school late year but I am pretty sure they are Edexcel restricted access ones for teachers only - so unfortunately for you there won’t be a mark scheme available. Sorry I can’t help further but best of luck with your exams!

Reply 19

Original post
by wbanner2001
Hi, I left school late year but I am pretty sure they are Edexcel restricted access ones for teachers only - so unfortunately for you there won’t be a mark scheme available. Sorry I can’t help further but best of luck with your exams!

Ah, that makes sense. Thank you anyway!

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