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Fp1 help

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
Reply 1
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Zacken
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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


Let u=z2u=z^2 then use your answer to the first part and back-substitute.
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.
Reply 3
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Super199
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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
Let u=z2u=z^2 then use your answer to the first part and back-substitute.

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
(edited 8 years ago)
Reply 4
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Zacken
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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 z=±−3+4i,±−3−4iz = \pm \sqrt{-3 + 4i}, \pm \sqrt{-3-4i}. You might want to quickly good how to find the square root of a complex number. Have you learnt De Moivre's theorem?
Reply 5
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Super199
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Original post by Zacken
Yes, that's the entire point. So now your rotos are given by z=±−3+4i,±−3−4iz = \pm \sqrt{-3 + 4i}, \pm \sqrt{-3-4i}. 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?
Reply 6
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Zacken
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Original post by Super199
Yh sort of but does that involve putting it into cos theta+isin theta form?


Yeah.
Reply 7
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Super199
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Original post by Zacken
Yeah.


Yh sorted cheers :smile:
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?
Reply 8
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Zacken
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Original post by Super199
Yh sorted cheers :smile:
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.
Reply 9
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Super199
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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 :smile:
Reply 10
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Zacken
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Original post by Super199
Yh got it cheers :smile:


Good work!

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