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July 30th, 2016, 07:13 AM   #1
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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!
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July 30th, 2016, 02:05 PM   #2
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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.
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July 30th, 2016, 02:34 PM   #3
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$\displaystyle z=w+3i+2\implies z-3i=w+2$

$\displaystyle z^2-iw+5-2i=z^2-i(w+2)+5=z^2-i(z-3i)+5=z^2-iz+2=0$
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July 31st, 2016, 07:29 AM   #4
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Quote:
Originally Posted by mathman View Post
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.
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July 31st, 2016, 07:33 AM   #5
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle z=w+3i+2\implies z-3i=w+2$

$\displaystyle z^2-iw+5-2i=z^2-i(w+2)+5=z^2-i(z-3i)+5=z^2-iz+2=0$
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
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July 31st, 2016, 03:07 PM   #6
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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 $w$.
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July 31st, 2016, 04:25 PM   #7
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$\displaystyle z^2-iz+2=0\implies z=\dfrac{i\pm\sqrt{i^2-8}}{2}=\dfrac{i\pm3i}{2},\quad z\in\{-i,2i\}$

Now use a given equation to find $w$.
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August 1st, 2016, 05:49 AM   #8
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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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