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July 30th, 2016, 07:13 AM  #1 
Newbie Joined: Apr 2016 From: Wonderland Posts: 16 Thanks: 0  Simultaneous complex equations
Hi all, I have spent a couple of hours on this perplexing question. Solve the simultaneous equations: z = w + 3i + 2 and z^2  iw + 5  2i = 0 giving z and w in the form (x + yi) where x and y are real. I tried various methods, all to no avail. I have substituted z into z^2, I got the wrong answers. I also tried letting z be (a + bi) and w be (c + di) and tried to combine the 2 equations together, and I got a horrible mess with many unknowns. Please help me, thank you! 
July 30th, 2016, 02:05 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,586 Thanks: 612 
Substituting (w+3i+2) for z in the second equation gives $\displaystyle w^2+2w(3i+2)5+6iiw+52i=0$. Collect terms to get $\displaystyle w^2+(4+5i)w+4i=0$. This looks like a straightforward (although messy) quadratic for w. 
July 30th, 2016, 02:34 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,859 Thanks: 1080 Math Focus: Elementary mathematics and beyond 
$\displaystyle z=w+3i+2\implies z3i=w+2$ $\displaystyle z^2iw+52i=z^2i(w+2)+5=z^2i(z3i)+5=z^2iz+2=0$ 
July 31st, 2016, 07:29 AM  #4 
Newbie Joined: Apr 2016 From: Wonderland Posts: 16 Thanks: 0  From your equation, I got w = [(45i) + √(9 + 24i)] / 2 or [(45i)  √(9 + 24i)] / 2 as the solutions but I am unable to solve from here.

July 31st, 2016, 07:33 AM  #5 
Newbie Joined: Apr 2016 From: Wonderland Posts: 16 Thanks: 0  I did manage to get $\displaystyle z^2iz+2=0$ from one of my tries, and from there I let z be (a + bi) and I expanded to obtain a quadratic with a couple of unknowns. I calculated a = 0 and b = 0.5 which is wrong 
July 31st, 2016, 03:07 PM  #6 
Math Team Joined: Nov 2014 From: Australia Posts: 688 Thanks: 243 
Do you know how to calculate the square root of a complex number? If you can do that, then you can just use the quadratic formula to find $z$ or $w$.

July 31st, 2016, 04:25 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,859 Thanks: 1080 Math Focus: Elementary mathematics and beyond 
$\displaystyle z^2iz+2=0\implies z=\dfrac{i\pm\sqrt{i^28}}{2}=\dfrac{i\pm3i}{2},\quad z\in\{i,2i\}$ Now use a given equation to find $w$. 
August 1st, 2016, 05:49 AM  #8 
Newbie Joined: Apr 2016 From: Wonderland Posts: 16 Thanks: 0 
Hi everyone, I finally managed to solve this question! Thank you all for your help =) I managed to use quadratic formula on the equation given by greg and solved for w and z. 

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complex, equations, simultaneous 
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