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 Complex Analysis Complex Analysis Math Forum

 July 30th, 2016, 07:13 AM #1 Newbie   Joined: Apr 2016 From: Wonderland Posts: 23 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: 6,704 Thanks: 669 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, 02:34 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 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, 07:29 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, 07:33 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, 03:07 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, 04:25 PM #7 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 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, 05:49 AM #8 Newbie   Joined: Apr 2016 From: Wonderland Posts: 23 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. Tags complex, equations, simultaneous Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post srahman33 Algebra 2 August 28th, 2014 02:55 AM jaz123 Algebra 2 March 13th, 2014 10:55 AM dralay Algebra 3 July 25th, 2013 02:47 AM queenie_n Complex Analysis 5 October 17th, 2012 05:20 PM bilano99 Algebra 3 June 30th, 2012 05:48 AM

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