Results 1 to 8 of 8

Thread: elp with complex numbers.

  1. #1
    Junior Member
    Joined
    May 2018
    From
    gol
    Posts
    29

    elp with complex numbers.

    Hello, I would be happy to help with complex numbers:
    Find equation solutions:
    elp with complex numbers.-20180518_173318.jpg
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,042
    Thanks
    2957
    Awards
    1

    Re: elp with complex numbers.

    Quote Originally Posted by yossa View Post
    Hello, I would be happy to help with complex numbers:
    Find equation solutions:
    Click image for larger version. 

Name:	20180518_173318.jpg 
Views:	3 
Size:	390.4 KB 
ID:	38745
    Sorry but there is no standard way to do these.
    Have a look at THIS SOLUTION.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    868
    Thanks
    409

    Re: elp with complex numbers.

    $\displaystyle z^3=-3 i \overset{-}{z}$

    Take absolute value

    $\displaystyle \left|z|^3=\text{3$|$z$|$}\right.$

    $\displaystyle |z|=0 \text{or} \sqrt{3}$

    $\displaystyle z=0$ is a solution

    To find the other solutions multiply the original equation by $\displaystyle z$

    $\displaystyle z^4=-9i$

    This can be solved to give four more solutions
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,981
    Thanks
    3121

    Re: elp with complex numbers.

    Write z= a+ bi with a and b real. Then $\displaystyle z^3= (a+ bi)^3= a^3+ 3a^2bi+ 3a(bi)^2+ (bi)^3= a^3+ 3a^2bi- 3ab^2- b^3i= (a^3- 3ab^2)+ (3a^3b- b^3)i$. The right side of the equation is $\displaystyle -3\overline{z}= -3(a- bi)= -3a+ 3bi$. Equating real and imaginary parts, $\displaystyle a^3- 3ab^2= -3a$ and $\displaystyle 3a^2b- b^3= 3b$. One obvious solution is a= b= 0 so z= 0. Also if a= 0, the second equation becomes $\displaystyle -b^3= 3b$ and if b is not 0, $\displaystyle -b^2= 3$ which, since b is real, is impossible. If b= 0, the first equation becomes $\displaystyle a^3= -3a$ and if a is not 0, $\displaystyle a^2= -3$, again, impossible. That is, z= 0 is a solution and all other solutions must have both a and b non-zero.

    If a and b are non-zero, we can divide $\displaystyle a^3- 3ab^2= -3a$ by a to get $\displaystyle a^2- 3b^2= -3$ and we can divide $\displaystyle 3a^2b- b^3= 3b$ by b to get $\displaystyle 3a^2- b^2= 3$. Adding those two equations, $\displaystyle 4a^3- 4b^3= 0$ so $\displaystyle a^3- b^3= (a- b)(a^2+ ab+ b^2)= 0$. If $\displaystyle a- b= 0$ then $\displaystyle a= b$ and we can write $\displaystyle 3a^2- b^2= 3$ as $\displaystyle 3a^2- b^2= 2a^2= 3$. $\displaystyle a^2= \frac{3}{2}$ so $\displaystyle a= \pm\sqrt{\frac{3}{2}}= \pm\frac{\sqrt{6}}{2}$. Two more solutions are $\displaystyle z= \frac{\sqrt{6}}{2}(1+ i)$ and $\displaystyle -\frac{\sqrt{6}}{2}(1+ i)$.

    If $\displaystyle a\ne b$ then we must have $\displaystyle a^2+ ab+ b^2= 0$. By the quadratic formula, $\displaystyle a= \frac{-b\pm\sqrt{b^2- 4b^2}}{2}= \left(\frac{-1\pm\sqrt{-3}}{2}\right)b= \left(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\right)b$. But that is impossible because a and b are both real.

    So the solutions to this equation are $\displaystyle z= 0$, $\displaystyle z= \frac{\sqrt{6}}{2}(1+ i)$, and $\displaystyle z= -\frac{\sqrt{6}}{2}(1+ i)$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    868
    Thanks
    409

    Re: elp with complex numbers.

    Quote Originally Posted by HallsofIvy View Post

    The right side of the equation is $\displaystyle -3\overline{z}= -3(a- bi)= -3a+ 3bi$.
    the right side is $\displaystyle -3i \bar{z}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2018
    From
    gol
    Posts
    29

    Re: elp with complex numbers.

    Should not I first find the r and the angle?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,042
    Thanks
    2957
    Awards
    1

    Re: elp with complex numbers.

    Quote Originally Posted by yossa View Post
    Should not I first find the r and the angle?
    This is not that type of question.
    Changing to rectangular form, expand & set the real and imaginary parts equal to zero.
    Then solve, SEE HERE.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    868
    Thanks
    409

    Re: elp with complex numbers.

    Quote Originally Posted by yossa View Post
    Should not I first find the r and the angle?
    yes.

    Let $\displaystyle z=r e^{i \theta }$

    Take absolute value to show that $\displaystyle r^3=3r$ and so either $\displaystyle r=0$ or $\displaystyle r=\sqrt{3}$

    $\displaystyle e^{3i \theta }= e^{i 3\pi /2} e^{-i \theta }=e^{i(3\pi /2-\theta )}$

    and

    $\displaystyle 3 \theta =3\pi /2-\theta +2 k \pi $

    $\displaystyle \theta =\frac{3\pi }{8}+\frac{\pi }{2}k$

    where $\displaystyle k=0,1,2,3$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: Mar 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Aug 25th 2009, 11:22 AM
  4. Replies: 2
    Last Post: Feb 7th 2009, 06:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

/mathhelpforum @mathhelpforum