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 October 26th, 2013, 08:17 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 solve quadratic equation How to solve this? $z^{2}-3z-5i=3iz$ $z^{2}-3z-5i-3iz=0$ $z^{2}-(3+3i)z-5i=0$ and I calculate delta $\Delta=(3+3i)^2-4(1\cdot 5i)=9+18i-9+20i=38i$ how to proceed?
October 26th, 2013, 08:30 AM   #2
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Re: solve quadratic equation

Quote:
 Originally Posted by rayman and I calculate delta $\Delta=(3+3i)^2-4(1\cdot 5i)=9+18i-9+20i=38i$ how to proceed?
It should be:

$(3+3i)^2-4(1)(-5i)=9+18i-9+20i=38i$

Then:

$z= \frac{(3+3i) \pm \sqrt{38i}}{2}$

October 26th, 2013, 08:46 AM   #3
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Re: solve quadratic equation

Quote:
 Originally Posted by Pero It should be: $(3+3i)^2-4(1)(-5i)=9+18i-9+20i=38i$ Then: $z= \frac{(3+3i) \pm \sqrt{38i}}{2}$
Thank you, it was a typo. I have a question: don't we have to simplify it somehow? I am thinking of $\sqrt{38i}$ or is it the final answer?

 October 26th, 2013, 09:20 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: solve quadratic equation You want to express z as a + bi for real numbers a and b. To help you do this, note that: $\sqrt{i}= \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$
 October 26th, 2013, 10:52 AM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: solve quadratic equation There are, of course, two complex numbers whose square is i. And since the complex numbers cannot be "ordered", we cannot assert that "$\sqrt{i}$ is the positive one" as we do with real numbers. We must say that $\sqrt{i}= \frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i$.
 October 27th, 2013, 01:07 PM #6 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: solve quadratic equation Okay but $\sqrt{i}= \pm( \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{2}}i)$
 October 27th, 2013, 01:32 PM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: solve quadratic equation $$$\frac{1}{sqrt{2}} \ - \ \frac{1}{sqrt{2}}i$$^2 \= \ -i$

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