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$\ z^2 = i-1 \ $ Hey guys, I couldn't find a way to solve this problem. The question suggests I replace $\ z\ $ with $\ x+iy\ $ and then go from there, but I always end up having to complete the square and end up with a completely different answer to the back o the book. Can you please help? Thanks

Antonios
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  • $(x + iy)^2 = x^2 - y^2 + i2xy$ so $2xy = 1$ and $x^2 - y^2 = -1$ – Azlif Jan 07 '18 at 12:12
  • @Azlif yes this is called the method of identification and it works. – mathreadler Jan 07 '18 at 12:13
  • I've reached that far but it wants me to solve for z @Azlif – Antonios Jan 07 '18 at 12:14
  • Why don't you show what result you got and how, so that we can point out the error in your calculation? – Martin R Jan 07 '18 at 12:17
  • Well, I have $\ x^2 - y^2 = -1 \ $ and $\ x = 1/2y \ $. I substituted my value for x into the other equation and eventually ended up with $\ 1/4y^2 - y^2 = -1 \ $. Then I got a quartic of $\ 4y^4 +4y^2 - 1 = 0 \ $. So I completed the square like a quadratic and got $\ 4(a - 0.5)^2 = 1.25 \ $.This gave me a value for a as $\ a = sqrt(5/16) + 0.5 \ $. Did I complete the square wrong? Thanks – Antonios Jan 07 '18 at 12:27
  • $\ a = y^2 \ $ as I didn't want to do it as a quartic – Antonios Jan 07 '18 at 12:28
  • Show your working, your answer, and the answer from the book. – it's a hire car baby Jan 08 '18 at 13:10

2 Answers2

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$i-1 = \sqrt2e^{3i\pi/4}$, so $z = \pm \sqrt[4]2e^{3i\pi/8}$.

Kenny Lau
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let $$z=x+iy$$ then we get $$x^2-y^2+2xyi=i-1$$ so we get $$x^2-y^2+1=0$$ and $$2xy-1=0$$ Can you solve this?

Dr. Sonnhard Graubner
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