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$$iz^2 + 3z -1 -3i=0$$

I am a bit confused on approaching this problem initially. I believe that $a,b,c$ are meant to be real numbers in the quadratic equation here. So would I be using these as coefficients?:

$$a=i$$ $$b=3$$ $$c=-1-3i$$

And then progress as $$z_1, z_2 = \frac{-b \pm\sqrt{4ac-b^2}}{2a}$$

Or am I completely wrong?

amWhy
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user382540
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  • Use the famous formula with these values for the coefficients and simplify. – Rafa Budría Mar 31 '17 at 19:41
  • Not at all, you are not completely wrong, but rather quite correct. Now just substitute the values of a, b, c into the equation on the right, two get two roots to the equation: $z_1, z_2$ – amWhy Mar 31 '17 at 20:10
  • @user382540 A cleaner (and in my view certainly easier) approach can be found here: http://math.stackexchange.com/questions/1648235/solving-a-quadratic-complex-equation/1648255#1648255 It avoids the quadratic formula – imranfat Mar 31 '17 at 20:15
  • What you've written is correct. You need to know how to find things like $\pm\sqrt{3+4i},\text{ etc.} \qquad$ – Michael Hardy Mar 31 '17 at 21:01

2 Answers2

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Your approach is completely correct...though there are some mistakes in the formula: the quadratic formula works as fine over the complex as over the reals (or over any other field with characteristic $\;\neq0\,$):

$$\Delta:=b^2-4ac=9-4\cdot i\cdot(-1-3i)=9+4i(1+3i)=-3+4i$$

and thus

$$z_{1,2}=\frac{-3\pm\sqrt{-3+4i}}{2i}$$

Now, evaluate the square root of $\;-3+4i\;$ and proceed...

DonAntonio
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    Surely, you must have known that this problem, or one such problem about a quadratic with complex coefficients, has been asked here. And a simple glance to the right of this very question does reveal such a question, reveals such a duplicate question with three answers. So why provide an answer to a duplicate question? If you're proud of your answer, post it at the better post of which this is a duplicate. You need to start thinking about how your actions may lead askers to not bother with searching for a similar question before posting, cuz folks like you will answer it anyway. – amWhy Mar 31 '17 at 19:55
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    @amWhy After searching almost desperately for many questions in this site along the years I don't try anymore. I think this site's "search" engine, or whatever, is very poor and incomplete and I don't care anymore. I already wrote about this somewhere several years ago and I don't care anymore. If I remember something I link it to the OP, otherwise I'd rather write an answer than waste my time searching for it. – DonAntonio Mar 31 '17 at 20:15
  • I'm not a fanatic about searching thoroughly across this site before answering, (the "search" in MSE is poor), but when I take a 10 second glance to the questions listed under "Related" on the right side of the webpages of questions, and see a question of a very similar sort, I think it is reasonable to first check the question out, before answering. That's my only point. Note, I did not downvote your answer. I'm sure I've inadvertently answered questions which turn out to be duplicates of other questions. So I don't mean to sound as though I'm "scolding" you. – amWhy Mar 31 '17 at 20:26
  • @imranfat Indeed so, and that problem with the search here is because it is "too free": you can put any title to your question. If there were only a few fixed, compulsory titles, and then some few subtitles, I think search would be greatly enhanced. Yet this, as many other things that've been brought up along the years, won't likely be changed. – DonAntonio Mar 31 '17 at 20:27
  • @imranfat I don't know what happened to your comment, but with respect to this question not being an exact duplicate, I disagree. If you haven't heard about abstract duplicates, search for "abstract duplicate(x)" on meta.math. Any question regarding how to solve a quadradic equation with complex coefficients is an exact duplicate to all these duplicate question. The coefficients a, b, c, exactly, do not matter, so long as one or two or three are complex. – amWhy Mar 31 '17 at 20:29
  • I'd have said something about how to evaluate things like $\pm\sqrt{-3+4i}.$ There I would use the tangent half-angle formula: $$ \tan \frac \theta 2 = \frac{\sin\theta}{1 + \cos\theta} = \frac{4/5}{1+(-3/5)} = \frac 4 {5-3} = 2 \text{ since } 5 = |-3+4i|. $$ – Michael Hardy Mar 31 '17 at 21:03
  • There I would use the tangent half-angle formula: $$ \tan \frac \theta 2 = \frac{\sin\theta}{1 + \cos\theta} = \frac{4/5}{1+(-3/5)} = \frac 4 {5-3} = 2 \text{ since } 5 = |-3+4i|. $$ So you get $ \left( \dfrac{1+2i} \bullet \right) $ where $\bullet$ is so chosen as to make the norm equal to $\sqrt 5.$ (And in this case that means $(\bullet=1).\qquad$ – Michael Hardy Mar 31 '17 at 21:09
  • A MathJax peculiarity $$ 5= |-3+4i|, \qquad 5=|{-3}+4i| \qquad 5=\left|-3+4i\right| $$ Notice how the first of these looks different from the others, and wrong, and notice how they're coded. It makes sense that the second and third ways of coding this would avoid the problem. Whether the first way should be expected to avoid the problem might bear discussion, but as you see, it does not avoid it. – Michael Hardy Mar 31 '17 at 21:13
2

The quadratic formula works even when the coefficients of the polynomial are complex numbers. In your case, we have $az^2 + bz + c = 0$ where

$$ a = i, b = 3, c = -1 - 3i $$

so the roots are given by

$$ z_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = -\frac{i}{2} \left( -3 \pm \sqrt{9 - 4i(-1 - 3i)} \right) = -\frac{i}{2} \left( -3 \pm \sqrt{-3 + 4i} \right).$$

All that is left to do is to find the (complex) square roots of $-3 + 4i$ and plug it into the formula.

levap
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