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October 20th, 2016, 07:37 AM  #1 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 47 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 08:55 PM. 
August 7th, 2017, 09:00 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395 
Hint: z³ + 3z² + 3z = z implies z(z² + 3z + 2) = 0, so z(z + 1)(z + 2) = 0.

August 8th, 2017, 04:31 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707  Quote:
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 10:12 PM.  
August 8th, 2017, 04:35 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,595 Thanks: 938 Math Focus: Elementary mathematics and beyond 
It seems our [latex]...[/latex] tags are no longer functioning. Use [math]...[/math] for $\displaystyle \LaTeX$.

August 8th, 2017, 10:35 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395 
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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