complex number (1 Viewer)

[samuelclarke]

find using De Moivre's theorem the complex numbers z which satisfy th equation z^5= z (conjugate)

 

[umm what]

Question is not clear :S

find using De Moivre's theorem the complex numbers z which satisfy th equation z^5= z (conjugate)

 

[lolcakes52]

he means (x+iy)^5=(x-iy) find x+iy.

 

[umm what]

what a sick question...lol cbb doing it
its not even hsc style question

he means (x+iy)^5=(x-iy) find x+iy.

 

[jyu]

Let z=rcis@, z^6=|z|^2, r^6 cis6@=r^2, cos6@ + isin6@=1/r^4, cos6@=1/r^4 and sin6@=0
.: 6@=2npi, @=npi/3, r=1
.: z=cis(npi/3) where n=0, +/- 1, +/- 2, 3.
i.e. z=1, 1/2+irt3 /2, 1/2-irt3 /2, -1/2+irt3 /2, -1/2-irt3 /2, -1

 

[seanieg89]

The correct answer is:

 

[Trebla]

A slightly different approach....After converting z to general mod-arg form, we have for integer k:



By equating the modulus, obviously r = 1 (since r is real and positive by definition of a modulus). Now equating the arguments:



which gives us the solution set for z

 

[Carrotsticks]

What about the trivial case z=0?

 

[bleakarcher]

The correct answer is:

Just a little question. By the fundamental theorem of algebra, there are only meant to be 5 solutions. How is it there are 6? Is it because the polynomial isn't purely in z but in conjugate of z?

 

[seanieg89]

Yeah its not a polynomial equation in z...