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I tried to solve this complex polynomial:

$$(1 + i)z^2 + (-1 + 7i)z - (10 - 2i) = 0$$

When I did the whole discriminant thing and I got $\sqrt{18i}$ which I don't know how to deal with. Usually my dicriminant doesn't contain the i inside of it. How are equations like these solved?

Thank you.

Fakemistake
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0_jump
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  • Try $(\frac{1+i}{\sqrt2})^2$ – almagest Jan 05 '20 at 17:17
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    The usual quadratic formula works if you get the square roots right. – Ross Millikan Jan 05 '20 at 17:18
  • In less obvious cases use the polar form, ie write the complex number as $re^{i\theta}$. Then the square roots are obvious. – almagest Jan 05 '20 at 17:20
  • I got the square doot of 18i with the standard formula but I don't feel like putting an i inside of a square root is right – 0_jump Jan 05 '20 at 17:23
  • https://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – Angina Seng Jan 05 '20 at 17:31
  • There’s nothing wrong with having a complex number under the radical sign, as long as you recognize that you’re dealing with an ill-defined construct (“two-valued function”, as people used to say). When the coefficients are not real, though, the only significant fact about the discriminant is whether it’s zero or not. – Lubin Jan 05 '20 at 17:45
  • For the people saying that this question is answered elsewhere I think you are right in a way, but I found Carot's answer to be really simple and immediately applicable. If the community deems this post too close to others than that's also okay if they want to remove it. – 0_jump Jan 05 '20 at 17:46
  • If you think a duplicate answers your question, you should close it as a dupe yourself. This is letting other people know that they can find additional information which may answer their question when they look for an answer here. – Simply Beautiful Art Jan 06 '20 at 02:41

3 Answers3

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Hint: Notice that $$18i = 9\cdot 2i = 9 \cdot (1+i)^2$$

nonuser
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  • I like this. Is it the standard way to approach this kind of problem (find a way to get rid of the i in side the square root) or is there a separate algorithm to go through in situations like this (instead of the discriminant thing)? – 0_jump Jan 05 '20 at 17:21
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You have to look for the square roots of $18i$ In the complex plane. The most common way of doing so is to put your complex number in polar form, here $18i=18 e^{i\pi/2}$. Then look for square roots of $18$ (should be easy) and $e^{i\pi/2}$.

Carot
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  • Note that $e^{i\theta/2}$ is a sqare root of $e^{i\theta}$. – Carot Jan 05 '20 at 17:23
  • In the present case, this method is easy. However, in the general case, there is a faster method, based on the classical problem of finding two real numbers, given their sum and their product. – Bernard Jan 05 '20 at 17:40
  • Well, yes and no. The question is what $\sqrt{2i}$ might be. For those who are experienced in dealing with Gaussian numbers, there’s no question at all: $(1+i)^2=2i$, as you verify easily. Just don’t forget about $-1-i$. – Lubin Jan 05 '20 at 17:47
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There is nothing at all wrong with representing the solutions by

$$x = \frac{(1 - 7i) + \sqrt{18i}}{2(1 + i)}$$

That is a perfectly valid, explicit mathematical expression on the right and if $x$ satisfies this equation, it satisfies the other. The trick is that, as mentioned in the comments, $\sqrt{a}$ is ambiguous for complex $a$-input - technically it is too for reals, but "less" so in that it is easier to choose between the two options. But as long as you understand that symbol as being substitutible for either of two values it represents, or you set up a convention ahead of time, then there is no problem here.

The_Sympathizer
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