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It concerns with finding roots of complex polynomial: $x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$

But, this is a guess game for me, with no formal process to get in this case.

The logical process is to take roots of the quadratic equation, but I hope the quadratic formula does not work for complex coefficients.

Update -- Based on comments by @WillJagy, the square root is $\pm1\pm3i$, hence the equation is having roots as :$\frac{-(i-1)\pm\sqrt{-6i-8}}2\implies \frac{(1-i)\pm(\pm 1\pm 3i)}2$ can have possible values:
(i) for $1+3i$, get: $x_{11} = 1+i, x_{12} = -2i$
(ii) for $1-3i$, get: $x_{21} = 1-2i, x_{22} = i$
(iii) for $-1+3i$, get: $x_{11} = i, x_{21} = 1-2i$
(iv) for $-1-3i$, get: $x_{11} = -2i, x_{21} = 1+i$

So, the $4$ root pairs should be tried one by one:

(i) $1+3i =>(x-1-i)(x+2i) = x^2 +ix -x -2i +2$, mismatch
(ii) $1-3i => (x-i)(x-1+2i) = x^2 -x +ix +i +2$, matches original equation
(iii) same as for (ii),
(iv) same as for (i)

Why only two root pairs match, and the other don't?

Some answers on mse : 1

Update 2 -- To add to the answer by @Skip, where the fourth quadrant angle was transformed to the first one by taking out $\sqrt{-1} = i$ as common factor. For mixed sign discriminant, let $D = -6 +8i$, then angle cannot be taken\manipulated to be in the first quadrant; & can lie in the 2nd or 3rd quadrant.
In 2nd quad., $\sin$ of negative angle is positive, while in 3rd quad. both $\sin, \cos$ are negative.

jiten
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    Your hopes will not be answered, because the quadratic formula does work for complex coefficients. –  May 27 '18 at 01:15
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    The quadratic formula works for complex coefficients – mallan May 27 '18 at 01:16
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    @Skip It is not working, as can be seen by applying that. $$x = \frac{-(i-1)\pm\sqrt{(i-1)^2-4(2+i)}}2\implies \frac{-(i-1)\pm\sqrt{-6i-8}}2$$ – jiten May 27 '18 at 01:19
  • Follow this link. If you crank a single bit more after that, I am off. –  May 27 '18 at 01:24
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    "It is not working". Be humble. It is not the formula's fault. It is you who doesn't know, it is you who doesn't understand. So, instead of blaming the formula, say "I don't know how to do it", "I don't understand". –  May 27 '18 at 01:29
  • @WillJagy The few possibilities by me (in integers) are : $g=\pm 1, h=\pm 3; g=\pm 3, h =\pm 1$. I do not know to get irrational roots, if it fits. – jiten May 27 '18 at 01:34
  • @WillJagy Thanks, but I feel that irrational roots should be possible also. How can they be found too? Also, for finding complex pair $g,h$, a formal approach should be easy, say (please correct, if needed) : $(a+bi)^2 = \sqrt{c^2+d^2}$, with $c=8, d=6$. – jiten May 27 '18 at 01:42
  • @WillJagy, the square root is $\pm1\pm3i$, hence the equation is having roots as :$\frac{-(i-1)\pm\sqrt{-6i-8}}2\implies \frac{(1-i)\pm(\pm 1\pm 3i)}2$ can have possible values: (i) for $1+3i$, get: $x_{11} = 1+i, x_{12} = -2i$; (ii) for $1-3i$, get: $x_{21} = 1-2i, x_{22} = i$; (iii) for $-1+3i$, get: $x_{11} = i, x_{21} = 1-2i$; (iv) for $-1-3i$, get: $x_{11} = -2i, x_{21} = 1+i.$ Only, two root pairs out of four suits, i.e. when roots are $(i, 1-2i), (1-2i, i)$, as shown in 'Update' above for cases (ii), (iii). Why? – jiten May 27 '18 at 02:18

2 Answers2

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$$x^2+(i-1)x+2+i=0$$

So

$$x=\frac{-(i-1)\pm\sqrt{-8-6i}}{2}=\frac{-(i-1)\pm i\sqrt{8+6i}}{2}$$

$$\sqrt{8+6i}=\sqrt{10}\left(\cos\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)\right)\space\space\space\text{k=0, 1}$$

Take $k=0$ since if $k=1$, we'll just get the negative of when $k=0$ and the $\pm$ in front of the square root takes care of this for us

These are nonstandard trig arguments but

$$\cos\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=\frac{3}{\sqrt{10}}$$

$$i\sin\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=i\frac{1}{\sqrt{10}}$$

Therefore

$$\sqrt{8+6i}=3+i$$

So

$$x=\frac{1-i\pm (-1+3i)}{2}$$

$$\boxed{x=i}$$

$$\boxed{x=1-2i}$$

Quadratic formula works

I'll keep the above because that is what appeared in the original problem, but here's how to evaluate $\sqrt{-6+8i}$.

$-6+8i$ is in the second quadrant

$$\theta=\tan^{-1}\left(\frac{8}{-6}\right)$$

But this angle describes a complex number in the fourth quadrant since $tan^{-1}(x)$ has the range $-\frac{\pi}{2}< x<\frac{\pi}{2}$, so add $\pi$ to swing it to the second quadrant

$$\theta=\tan^{-1}\left(\frac{8}{-6}\right)+\pi$$

Its magnitude is still $10$, so for $k=0,\space 1$

$$\sqrt{-6+8i}=\sqrt{10}\left(\cos\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi+2\pi k}{2}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi+2\pi k}{2}\right)\right)$$

These arguments are not standard by any means, so I admittedly used a calculator and will again for evaluating this one too, which is why there is no work shown in evaluating them. The point of this post though is that the quadratic formula works for complex numbers.

I like Siong's way for solving the square root; the results will be the same and it's nicer to work with in this case. Sometimes this way is easier in case the polynomial isn't easily factorable, it depends on the problem

Taking $k=0$

$$\cos\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi}{2}\right)=\frac{1}{\sqrt{5}}$$

$$i\sin\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi}{2}\right)=i\frac{2}{\sqrt{5}}$$

So

$$\sqrt{-8+6i}=\sqrt{2}(1+2i)$$

Taking $k=1$ will result in $\sqrt{-8+6i}=\sqrt{2}(-1-2i)$ but the plus and minus takes care of this for us

$$\pm\sqrt{2}(1+2i)=\mp\sqrt{2}(-1-2i)$$

Say if the radicand were $6-8i$, then this lies in the fourth quadrant, so we would not have added $\pi$ when finding its angle.

mallan
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  • Please help with my doubt in 'Update' to OP(post), i.e. why only two values of square root : $\pm 1 \pm 3i$ suit? Your answer also concerns with it, but request some more elaboration. – jiten May 27 '18 at 02:22
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    Square roots only give 2 solutions. 4 solutions from a single square root is out of the question. There is a slight discrepancy in how we each handled the radicand; the 4 answers came from solutions to 2 different square roots. Notice I factored out $-1$ to flip the complex number into the first quadrant from the third. This changed its angle by $-\pi$.

    The solutions to $\sqrt{-8-6i}$ are {$1-3i, -1+3i$} whereas solutions to $\sqrt{8+6i}$ are {$3+i, -3-i$}, but I factored out $i$ to make that form, so $i${$3+i,-3-i$}$=${$-1+3i,1-3i$}. Therefore, they are the same, only 2 solutions

     – mallan May 27 '18 at 03:07
  • Yes, by deftness you have made the trigonometrical soln. to lie in the first quadrant (by factoring out $i$, so as to remove -ve sign). But, if the discriminant were of mixed sign, say : $-8+6i$, then it confuses me. Can you please modify your soln. to take that as discriminant. The real challenge for me lies when the discriminant cannot be modified to take the angle in the first quadrant. – jiten May 27 '18 at 07:56
  • I am unable to understand how $$\cos\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=\frac{3}{\sqrt{10}}$$ & similarly for $\sin$ expression. – jiten May 27 '18 at 10:27
  • Thanks a lot. Your way is trigonometrical & is tougher to algebraic approach, i.e. what I feel. You have stated that it depends on problem, which of two approaches is easier. I request some good (tough) case for showing when trig. app. is easier; as feel it is tougher. Also, have confusion over stating magnitude, in answer for either $8+6i$ or $-6+8i$, as $\sqrt{10}$, when have stated it to be $10$. 3rd, would still request to show calculations for: $\cos\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=\frac{3}{\sqrt{10}}$, as would not expect $\sqrt{10}$ as denominator by calculator. – jiten May 27 '18 at 19:19
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    Right, call $a+bi=z$ so $|z|=\sqrt{a^2+b^2}$ and $z=|z|e^{\tan^{-1}(b/a)+2\pi k}$, $k\in\mathbb{Z}$, so $\sqrt{z}=\sqrt{|z|}e^{(\tan^{-1}(b/a)+2\pi k)/2}$. This approach is generalized for all forms $z^{1/n}$ including more complicated expressions like obtaining all values for $(2-i)^{2+i}$. I calculated nothing for the $\cos$ and $\sin$, I only typed them into Wolfram – mallan May 27 '18 at 19:28
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Let $z=a+bi, a,b, \in \mathbb{R}$, where $$z^2 = -8-6i$$

$$(a+bi)^2=(-8-6i)$$

$$a^2-b^2+2abi=-8-6i$$

that is we have

$$a^2-b^2=-8\tag{1}$$

$$ab=-3\tag{2}$$

Substituting $(2)$ into $(1)$,

$$a^2-\frac{9}{a^2}=-8$$

$$a^4+8a^2-9=0$$

$$(a^2-9)(a^2+1)=0$$

$$a=\pm3$$

when $a=3$, from $ab=-3$,$b=-1$.

when $a=-3$, from $ab=-3$, $b=1$

There are only two solutions for $z^2=-8-6i$, $z_1 = 3-i$ and $z_2 = -3+i$ and we have $z_2 = -z_1$. That is the solutions to $z^2 = -8-6i$ are $\pm(3-i)$

The solution to the original problem is

$$x=\frac{(1-i) \pm z_1}{2}$$

You might have solved for $a$ and substitute it into $(1)$ and thought that for each $a$, there are $2$ possible values of $b$'s. However, remember that you have to check all the constraitns, constraint $(2)$ tells us that for each non-zero $a$, there is exactly one $b$.

You shold be expecting $2$ roots counting multiplicity for a quadratic equation from fundamental theorem of algebra.

Siong Thye Goh
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