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 July 30th, 2016, 08:13 AM #1 Newbie   Joined: Apr 2016 From: Wonderland Posts: 18 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, 03:05 PM #2 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 Substituting (w+3i+2) for z in the second equation gives $\displaystyle w^2+2w(3i+2)-5+6i-iw+5-2i=0$. 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, 03:34 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond $\displaystyle z=w+3i+2\implies z-3i=w+2$ $\displaystyle z^2-iw+5-2i=z^2-i(w+2)+5=z^2-i(z-3i)+5=z^2-iz+2=0$
July 31st, 2016, 08:29 AM   #4
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 Originally Posted by mathman Substituting (w+3i+2) for z in the second equation gives $\displaystyle w^2+2w(3i+2)-5+6i-iw+5-2i=0$. Collect terms to get $\displaystyle w^2+(4+5i)w+4i=0$. This looks like a straightforward (although messy) quadratic for w.
From your equation, I got w = [(-4-5i) + √(-9 + 24i)] / 2 or [(-4-5i) - √(-9 + 24i)] / 2 as the solutions but I am unable to solve from here.

July 31st, 2016, 08:33 AM   #5
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 Originally Posted by greg1313 $\displaystyle z=w+3i+2\implies z-3i=w+2$ $\displaystyle z^2-iw+5-2i=z^2-i(w+2)+5=z^2-i(z-3i)+5=z^2-iz+2=0$
I did manage to get $\displaystyle z^2-iz+2=0$ 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, 04:07 PM #6 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 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, 05:25 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond $\displaystyle z^2-iz+2=0\implies z=\dfrac{i\pm\sqrt{i^2-8}}{2}=\dfrac{i\pm3i}{2},\quad z\in\{-i,2i\}$ Now use a given equation to find $w$.
 August 1st, 2016, 06:49 AM #8 Newbie   Joined: Apr 2016 From: Wonderland Posts: 18 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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