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# Fp1 help watch

1. Need help with 7b. Not too sure what to do. I did part a by completing the square but in guessing I use my answer to this part some how.
Cheers
http://m.imgur.com/6eU7OFZ
2. (Original post by Super199)
Need help with 7b. Not too sure what to do. I did part a by completing the square but in guessing I use my answer to this part some how.
Cheers
http://m.imgur.com/6eU7OFZ
3. Make a substitution w=z^2.

Solve how you normally would solve a quadratic to find w.

Then use the w=z^2 to find z.
4. (Original post by poorform)
Make a substitution w=z^2.

Solve how you normally would solve a quadratic to find w.

Then use the w=z^2 to find z.
(Original post by Zacken)
But dont you just go back to the same quadratic as before.
-3+4i =z^2
-3-4i=z^2

Sqaure roots of those give me the other roots? I cant remember how you do that
Is this when you do (a+bi)^2=-3+4i
5. (Original post by Super199)
But dont you just go back to the same quadratic as before.
-3+4i =z^2
-3-4i=z^2

Sqaure roots of those give me the other roots? I cant remember how you do that
Yes, that's the entire point. So now your rotos are given by . You might want to quickly good how to find the square root of a complex number. Have you learnt De Moivre's theorem?
6. (Original post by Zacken)
Yes, that's the entire point. So now your rotos are given by . You might want to quickly good how to find the square root of a complex number. Have you learnt De Moivre's theorem?
Yh sort of but does that involve putting it into cos theta+isin theta form?
7. (Original post by Super199)
Yh sort of but does that involve putting it into cos theta+isin theta form?
Yeah.
8. (Original post by Zacken)
Yeah.
Yh sorted cheers
How do I do part d?
|z-z1| <2 isnt that a circle i cant remember tbh.
And the angle of z1 lies between 0 and pi/2?
9. (Original post by Super199)
Yh sorted cheers
How do I do part d?
|z-z1| <2 isnt that a circle i cant remember tbh.
And the angle of z1 lies between 0 and pi/2?
|z-z_1| < 2 is the inside of a circle centred around z_1 with radius 2. The angle of z_1 lying between 0 and pi/2 just lets you know which root to pick as only one of the roots will be in the first quadrant and that is z_1.
10. (Original post by Zacken)
|z-z_1| < 2 is the inside of a circle centred around z_1 with radius 2. The angle of z_1 lying between 0 and pi/2 just lets you know which root to pick as only one of the roots will be in the first quadrant and that is z_1.
Yh got it cheers
11. (Original post by Super199)
Yh got it cheers
Good work!

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