Question

Solve the equation:
z^3=z+z*
I'm super lost; been trying to expand using z=a+bi but without going anywhere. I also tried utilizing z^n=r^n(cos(alpha)+i*sin(alpha)) without not really knowing what to do. Please halp! :c

Answer 1

Lol im studying tat right now

Answer 2

Ill help you in this when I get a chance tomorrow, tag me in this so I can find it tomorrow

Answer 3

Just got this one question; I know the solutions. Got no clue how to get there though.

Answer 4

Does z* imply the conjugate?

Answer 5

Indeedio.

Answer 6

I was not sure how to write it out.

Answer 7

okay, i will try.

Answer 8

Thanks! :) Trying to come up with new stuff as we speak as well.

Answer 9

Try, Z= a+bi.
you get
\[(a+ib)^3 = 2a\]
Now expand.
The terms with 'i' add up to zero.
The rest equal to 2a.

Answer 10

How do they add up to zero?

Answer 11

I get a^3-2ab^2-ab^2-2a+(3a^2b-b^3)*i=0

Answer 12

Because the imaginary part on both the sides should be equal.
I will give the working in next post.

Answer 13

Oh, right!

Answer 14

\[a^3+3a^2bi-3ab^2-ib^3 =a+ib+a-ib \]
\[(a^3-3ab^2)+i(3a^2b-b^3)=2a\]
Now equating real and imaginary parts.
\[a^3-3ab^2=2a \]
\[3a^2-b^3=0\]
Can you continue from here?

Answer 15

This is precisely the step I'm at.

Answer 16

a=(a^3-3ab^2)/2 and insert into (2) in the set. and pray to god it gets better(?).

Answer 17

The last equation has b multiplied to a^2. sorry.
I feel this is better.
On simplifying second one,\[3a^2=b^2\]
Now put b^2 in the first.
You have a third degree equation in 'a'.

Answer 18

I'll manage from here, thanks a lot! :)

Answer 19

(I think!)

Answer 20

I get the fator a(4a^2+1)=0

Answer 21

Good, I got stuck up right there.

Answer 22

But solving (4a^2+1)=0 yields wrong answer for a. :/

Answer 23

a=+-i/2

Answer 24

Looks like there was nothing wrong in the procedure. But the result is wrong.

Answer 25

Indeed, maybe the method is not viable?

Answer 26

If we only had a(a^2-2)=0 for some reason. :x

Answer 27

ooooh.

Answer 28

Aren't you dividing with zero at 3a^2b=b^3 --> 3a^2=b^2

Answer 29

b may be zero.

Answer 30

Yeah. :/

Answer 31

If we solve using trigonometry, (try)
\[\cos 3\theta =2 \cos \theta \]
\[\sin 3\theta = 0\]

Answer 32

Where did you go from?

Answer 33

\[z= \cos \theta + i \sin \theta \]

Answer 34

How cos(3Theta)? I'm stupidly bad at trigonometry. :c

Answer 35

From de' moivre's theorem.

Answer 36

Isn't z^n=r^n*(cos(Theta)+i*sin(Theta))?

Answer 37

No,
\[(\cos \theta + i \sin \theta)^n = \cos n \theta +i \sin n \theta\]

Answer 38

Ok yeah, I follow to cos(3Theta)=2cos(Theta) now.

Answer 39

Okay.do we get a answer this way?

Answer 40

I got noooooo idea dude! :D Not sure where to go from cos(3Theta)=2cos(Theta) and sin(3*theta)=0

Answer 41

arcsin(0)=0 --> Theta=0 (?)

Answer 42

Which one is the theta?

Answer 43

I'm super confused and got no idea what's happening atm.

Answer 44

What are your thoughts? Have you even solved it perhaps?

Answer 45

I am again getting a absurd result that cos is grater than 1.
Before getting it I divided the equation by cos.

Answer 46

:C

Answer 47

I am not able to solve this. Not getting anything.
Is there any solution for this?

Answer 48

The solutions are z=0,z=-sqrt(2),z=sqrt(2)

Answer 49

z=0 is obvious. I'll try. If you get a way, inform me.

Answer 50

You know, the set of equations pan out correctly, perhaps they're solvable somehow?

Answer 51

a^3-3ab^2=2a,3a^2b=b^3
that is

Answer 52

Or perhaps that way is only solvable numerically.

Answer 53

Super conundrum. :/

Answer 54

Yeah,
Using guess method we get the answers this way.
\[b(3a^2-b^2) = 0\]
\[a(a^2-3b^2-2)=0\]
Let b=0 be the solution of first equation.
Put it in the second assuming a is not zero.
You get a = +-sqrt(2).

Answer 55

I don't deal in guesstimating. :D

Answer 56

The other possibilities are not possible, since we just now tried to do.(above)

Answer 57

This is very solvable, I just don't know how. :(

Answer 58

Try any other way, you will end up with a wrong answer.
This is the only possible way. I couldn't see it until now. :(

Answer 59

But you can't just assume a is not zero.

Answer 60

But you get a is sqrt(2). So, it confirms that your assumption is right. You can then verify using the question.

Answer 61

Of course it might be right, that doesn't make it right. As I said, this is 100% solvable without assumptions or numerical methods.

Answer 62

Man I feel dumb. :s

Answer 63

Okay, When you get the method, post it. As far as I know, we use either z=a+ib or the trigonometry. Keep trying..

Answer 64

I didn;t read all of the above, did you get get the answer ?

Answer 65

Nope!

Answer 66

I know the answer, I've known it before I even asked here.

Answer 67

It's the way to the answer I need. :/

Answer 68

what is the answer ?

Answer 69

z=0,z=sqrt(2),z=-sqrt(2)

Answer 70

this means 3a^2 = b^2 should be wrong.

Answer 71

its true only for z=0. hmm

Answer 72

Indeed it is(?).

Answer 73

Arriving at 3a^2=b^2 you divide with b which is potentially 0.

Answer 74

(a+ib)^3 = 2a
a^3 - b^3 i + 3ib a^2 - 3ab^2 = 2a
(a^3 - b^3 - 3ab^2 -2a) + i ( 3b a^2 - b^3) =0

so b(3a^2 - b^2) = 0

b=0 is 1 solution
plugging that in (a^3 - b^3 - 3ab^2 -2a) = 0, you get the required value of a.

Only questions remains why we ignored 3a^2 - b^2 =0
hmm

Answer 75

Isn't b=0 only part of a solution?

Answer 76

b=0 is a part of the solution.

Answer 77

what is z* ? conjugate of z ?

Answer 78

Yes! So sorry for the late reply, didn't think you would respond so quickly.

Answer 79

\[ z^3 = 2 \Re[z] \]
has to be something that if rotated by 3 times goes to real axis.
\[ \large r^3 e^{i3\theta }= r^3 e^{in \pi }\]

Answer 80

try solving this for other value than zero, might work but not sure

Answer 81

hmm

Answer 82

I'm not sure what you mean or what direction I'm supposed to be heading in.

Answer 83

0+n*2Pi=Theta gives r(r^2-2)=0
The solutions are supposed to be z=0,z=-sqrt(2),z=sqrt(2)

Answer 84

The solutions to z^3=z+z* that is.""

Answer 85

z=0 is trivial solution

Answer 86

Sure

Answer 87

\[ \theta = 0, 60^\circ, 120, ... \]
http://www.wolframalpha.com/input/?i=%28r%29%5E3+%3D+2+r+cos%2860+degree%29

\[ z = \pm e^{i 2\pi/ 6}\]

Answer 88

I'm sorry, I don't follow at all. :(

Answer 89

\[ z = 0, \frac 1 2 + i \frac 1 2, -(\frac 1 2 + i \frac 1 2) \]

Answer 90

But: http://www.wolframalpha.com/input/?i=Solve+z%5E3%3Dz%2Bconjugate%28z%29

Answer 91

woops!! where did it go wrong then

Answer 92

looks like i made some mistake
http://www.wolframalpha.com/input/?i=%28E%5E%28I+Pi%2F3%29%29%5E3+%3D+E%5E%28I+Pi%2F3%29+%2B+E%5E%28-I+Pi%2F3%29

Answer 93

This problem is killing me. (Thanks so much for trying to help me btw)
Got no clue what mistake you did.

Answer 94

it seems like \( \theta = 0 \) is the only solution
http://www.wolframalpha.com/input/?i=%28E%5E%28I+2Pi%2F3%29%29%5E3+%3D+E%5E%28I+2Pi%2F3%29+%2B+E%5E%28-I+2Pi%2F3%29

Answer 95

\[ r^3 = 2r \cos (0) = 2 r \]
it gives you 0, +- sqrt(2)

Answer 96

So that WAS it, god damn it problem. :D

Answer 97

But how would you draw that conclusion algebraically?

Answer 98

let \( z = r e^{i \theta }\) be your solution.
since z^3 = 2 Re z = 2 r cos(theta),
3 theta = npi => theta = n pi/3
show that n=0 is only solution and solve for 'r'