My Math Forum problem with solving the following equation

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 October 20th, 2016, 08:37 AM #1 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 problem with solving the following equation Hi everyone, I would be very grateful if anyone can help me solve this equation. I tried to solve it in many different ways, but I still have not managed to solve it. Please, help me! z^3 + 3z^2 + 3z = z (conjugation) Last edited by skipjack; August 7th, 2017 at 09:55 PM.
 August 7th, 2017, 10:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 Hint: z³ + 3z² + 3z = z implies z(z² + 3z + 2) = 0, so z(z + 1)(z + 2) = 0.
August 8th, 2017, 05:31 PM   #3
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 Originally Posted by srecko Hi everyone, I would be very grateful if anyone can help me solve this equation. I tried to solve it in many different ways, but I still have not managed to solve it. Please, help me! z^3 + 3z^2 + 3z = z (conjugation)
I presume you mean $\displaystyle z^3+ 3z^2+ 3z= \overline{z}$. That makes it a little harder than what skipjack says.

Taking $z= x+ iy$, $\displaystyle z^2= x^2- y^2+ 2ixy$ and $\displaystyle z^3= x^3- 3xy^2+ (3x^2y- y^3)i$

So $\displaystyle z^3+ 3z^2+ 3z= x^3- 3xy^2+ (3x^2y- y^3)i+ 3x^2- 3y^2+ 6ixy+ 3x+ 3iy= x^3- 2xy^2+ 3x^2- 3y^2+ 3x+ (3x^2y- y^3+ 6xy+ 3y)i$

Of course, $\displaystyle \overline{z}= x- iy$, so that $\displaystyle z^3+ 3z^2+ 3z= \overline{z}$ becomes $\displaystyle x^3- 2xy^2+ 3x^2- 3y^2+ 3x+ (3x^2y- y^3+ 6xy+ 3y)i= x- iy$ or $\displaystyle x^3- 2xy^2+ 3x^2- 3y^2+ 2x+ (3x^2y- y^3+ 6xy+ 4y)i= 0$

So we must have $\displaystyle x^3- 2xy^2+ 3x^2- 3y^2+ 2x= 0$ and $\displaystyle 3x^2y- y^3+ 6xy+ 4y= 0$.

The second equation has a $y$ in each term, so we can factor it out: $\displaystyle y(3x^2- y^2- 6x+ 4y)= 0$ so either $y= 0$ or $\displaystyle 3x^2- y^2- 6x+ 4y= 0$.

If $y= 0$, the first equation becomes $\displaystyle x^3+ 3x^3+ 2x= x(x^2+ 3x+ 2)= 0$. Either $x= 0$ or $\displaystyle x^2+ 3x+ 2= (x+ 2)(x+ 1)= 0$.

So ($x$, $y$) = (0, 0) or (-2, 0) or (-1, 0) are solutions.

The hard part of course, is if $y$ is not 0, so $\displaystyle 3x^2- y^2- 6x+ 4y= 0$. Completing the square, $\displaystyle (3x^2- 6x)- (y^2- 4y)= 3(x^2- 6x+ 9)- 27- (y^2- 4y+ 4)+ 4= 3(x- 3)^2- (y- 2)^2- 23= 0$ or $\displaystyle 3(x- 3)^3- (y- 2)^2= 23$.

Last edited by skipjack; August 8th, 2017 at 11:12 PM.

 August 8th, 2017, 05:35 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,662 Thanks: 965 Math Focus: Elementary mathematics and beyond It seems our $...$ tags are no longer functioning. Use $...$ for $\displaystyle \LaTeX$.
 August 8th, 2017, 11:35 PM #5 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 I gave a hint, not a full solution! It's not that hard, Country Boy. You made two errors in the initial algebra and then went badly wrong in the "hard part". However, there's a much easier solution method. As $(z + 1)^3 = \bar{z} + 1$ implies $(\bar{z} + 1)^3 = z + 1$, $(z + 1)^9 = z + 1$, which is easy to solve.

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