Complex Roots

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Homework Statement


Solve z^5 + 16 conjugate(z) = 0 for z element of C.

z^5 + 16z' = 0

http://puu.sh/2EBqC.png [Broken]


Homework Equations





The Attempt at a Solution



My first thought was to use z = a+bi and z' = a-bi
So:
(a+bi)5 + 16*(a-bi) = 0 + 0i And then expand and simplify to the real and non real parts. http://puu.sh/2EBwU.png [Broken]
But it seems WAAY too complicated for a question worth only 4 marks.
I would appreciate if someone could post their thoughts.
Thanks!


According to WolframAlpha: http://puu.sh/2EBKH.png [Broken]
 
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Answers and Replies

  • #2
tiny-tim
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hi planauts! :smile:

(try using the X2 button just above the Reply box :wink:)
Solve z^5 + 16 conjugate(z) = 0 for z element of C.
try multipying by z :wink:
 
  • #3
Dick
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Homework Statement


Solve z^5 + 16 conjugate(z) = 0 for z element of C.

z^5 + 16z' = 0

http://puu.sh/2EBqC.png [Broken]


Homework Equations





The Attempt at a Solution



My first thought was to use z = a+bi and z' = a-bi
So:
(a+bi)5 + 16*(a-bi) = 0 + 0i And then expand and simplify to the real and non real parts. http://puu.sh/2EBwU.png [Broken]
But it seems WAAY too complicated for a question worth only 4 marks.
I would appreciate if someone could post their thoughts.
Thanks!

According to WolframAlpha: http://puu.sh/2EBKH.png [Broken]
I think you'll have better luck with polar form than with rectangular. Try it.
 
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  • #4
BruceW
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I agree with these guys. Also, there are more solutions than wolfram alpha said. the button "more roots" gives the rest. (just in case you were confused that your roots were not the same as the ones you saw in wolfram alpha).
 
  • #5
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hi planauts! :smile:

(try using the X2 button just above the Reply box :wink:)

try multipying by z :wink:

[itex]z^6 + 16|z|^2 = 0[/itex]
[itex](z^3+4|z|i)(z^3-4|z|i) = 0[/itex]
Factor further using sum and differences of cubes?

I think you'll have better luck with polar form than with rectangular. Try it.
I'm not sure what you mean,

[itex]r^5e^{5i\theta}+16re^{-i\theta}=0[/itex]
[itex]r(r^4e^{5i\theta}+16e^{-i\theta})=0[/itex] r = 0
 
  • #6
Dick
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[itex]
I'm not sure what you mean,

[itex]r^5e^{5i\theta}+16re^{-i\theta}=0[/itex]
[itex]r(r^4e^{5i\theta}+16e^{-i\theta})=0[/itex] r = 0
That's exactly what I mean. Now multiply both sides by ##e^{i\theta}## and start thinking about what ##\theta## and r values might give you a solution.
 
  • #7
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That's exactly what I mean. Now multiply both sides by ##e^{i\theta}## and start thinking about what ##\theta## and r values might give you a solution.
I end up with.
[itex]r^4e^{6i\theta} = -16[/itex]

I know theta = -pi/2 and r = 2 would give a solution...

How would I manipulate this equation so I can use the De Moivre theorem?

Should I let 4w = 6theta

So
[itex]r^4*e^{4wi} = -16[/itex]

Let [itex]p^4 = r*e^{wi} = -16[/itex]
And then solve for p and change the w angles to theta? I know that r has to be definitely 2 though.

Am I over complicating things again?
 
  • #8
Dick
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I end up with.
[itex]r^4e^{6i\theta} = -16[/itex]

I know theta = -pi/2 and r = 2 would give a solution...

How would I manipulate this equation so I can use the De Moivre theorem?

Should I let 4w = 6theta

So
[itex]r^4*e^{4wi} = -16[/itex]

Let [itex]p^4 = r*e^{wi} = -16[/itex]
And then solve for p and change the w angles to theta? I know that r has to be definitely 2 though.

Am I over complicating things again?
Maybe. Yes, r must be 2. So ##e^{6i\theta}=(-1)##. That means ##cos(6\theta)=(-1)## and ##sin(6\theta)=0##, right? That's deMoivre. I don't think it should be hard from there.
 
  • #9
tiny-tim
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hi planauts! :smile:
[itex]z^6 + 16|z|^2 = 0[/itex]
[itex](z^3+4|z|i)(z^3-4|z|i) = 0[/itex] …
no, it's easier now to put z = |z|e instead,

giving you |z|4e6iθ = -24

(same as you got anyway)
 

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