My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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.
srecko is offline  
 
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.
skipjack is offline  
August 8th, 2017, 04:31 PM   #3
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,731
Thanks: 707

Quote:
Originally Posted by srecko View Post
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 10:12 PM.
Country Boy is offline  
August 8th, 2017, 04:35 PM   #4
Global Moderator
 
greg1313's Avatar
 
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 $\displaystyle \LaTeX$.
greg1313 is offline  
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.
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
complex, equation, problem, solving



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
solving an easy equation problem >_< DannyChu Algebra 3 January 28th, 2013 01:58 PM
solving equation problem CAMILLE Algebra 9 October 10th, 2008 05:36 AM
solving equation problem u_r_b_a_n Algebra 1 September 30th, 2008 01:51 AM
Problem solving an equation yaDDy Abstract Algebra 6 July 24th, 2008 12:22 PM
solving an easy equation problem >_< DannyChu Number Theory 1 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.