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Hi I tried to solve the equation $$ z^2 - (1+m)(1+i)z + i(m^2+1)=0 $$ but I don't know if my answer is wrong or right.

My first $\Delta$ was $-2(im^2+i-2im)$, the second one $0$. So $$ z_1 = (1+m)(1+i)+i\sqrt{im^2+i-2im} $$ Am I right or wrong please?

Bumblebee
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Amine Marzouki
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  • Where is the sign of equality? – Olimjon Jan 06 '16 at 06:08
  • @Olimjon Oh sorry it's equals to 0 – Amine Marzouki Jan 06 '16 at 06:11
  • What is $\Delta$? It is also generally not a good idea to take the square root (or any root of degree $> 1$) of a complex number, because (without doing some work) there is no good way to define a square root function on $\mathbb{C}$. – Jendrik Stelzner Jan 06 '16 at 06:16
  • Why not just set $z = a + bi$, expand, equate the real and imaginary parts, and solve a quadratic? Is there supposed to be a shortcut? – DanielV Jan 06 '16 at 06:17
  • @JendrikStelzner I really didn't get your meaning, why it's bad idea? it's the only way to solve the equations of the second degree, besides factoring of course ? – Amine Marzouki Jan 06 '16 at 06:21
  • @DanielV It'll be very long, and we need to discuss cases of m – Amine Marzouki Jan 06 '16 at 06:21
  • You forgot to divide $(1 + m)(1 + i)$ by $2$. Also, it seems to me that there is an error of a factor of $2$ under your square root sign. To find the square root of $\Delta$, it may help to write $\Delta = -2i(m^2 - 2m + 1)$. – David Jan 06 '16 at 06:27
  • @AmineMarzouki: While for a non-negative real number $x \geq 0$ one can define $\sqrt{x}$ as the unique non-negative root of the polynomial $p(t) = t^2-x$, this is approach does not work on the complex numbers $\mathbb{C}$, because they are not an ordered field. So it is generally a bad idea to write $\sqrt{z}$ with $z \in \mathbb{C}$, as long as you don’t have $z \in \mathbb{R}$ with $z \geq 0$. Instead try to work with squares of your numbers, or, if necessary, try to find an element $w \in \mathbb{C}$ with $w^2 = z$. Also, completing the square is your friend. – Jendrik Stelzner Jan 06 '16 at 06:35
  • But im not getting what you want to solve for ?? Is it $m$ – Archis Welankar Jan 06 '16 at 06:45
  • @ArchisWelankar no it's z – Amine Marzouki Jan 06 '16 at 06:46
  • @JendrikStelzner You mean it's not unique? – Amine Marzouki Jan 06 '16 at 06:47
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    $$c = (m^2 + 1)i = (m+i)(m-i)i = (-1 + mi)(m - i) = (m + i)(1+mi)$$ $$b = (1 + m)(1 + i) = 1 + m + (m + 1)i$$ since you want $z_1z_2 = c$ and $z_1 + z_2 = b$ then you can maybe guess that $z_1 = m+i$ and $z_2 = 1 + mi$. – DanielV Jan 06 '16 at 06:48
  • @DanielV Thanks I think that's it – Amine Marzouki Jan 06 '16 at 06:52
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    @AmineMarzouki Whether that's it depends on what method you are trying to learn. – DanielV Jan 06 '16 at 06:52
  • @AmineMarzouki: Yes, and unlike for the non-negative real numbers $\mathbb{R}_{\geq 0}$ we have no good way to solve this problem (without enough constraints and machinery). So when you write $\sqrt{z}$ for $z \in \mathbb{C}$ you refer to two possible values, without specifying which you mean. – Jendrik Stelzner Jan 06 '16 at 07:07
  • @DanielV +1 for depends on what method you are trying to learn. The original equation factorizes as $(z-b)(z-c)=0$ with your $b$, $c$ by elementary inspection, without a need to involve discriminants at all. But of course that doesn't generalize to less friendly exercises. – dxiv Jan 06 '16 at 07:56
  • http://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula/49243#49243 – Bumblebee Jan 06 '16 at 08:13

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By completing the square, we can obtain $$(z-(1+m)(1+i))^2=-\frac12 (5m^2+2m-1)~{\rm cis}~(\frac{3\pi}{2}+2k\pi)$$, where k is an integer. Then, we apply De'moivres theorem, which states that for some integer $n$, the set of solutions for $z^n=r\bigg(\cos(\theta+2k\pi) + i~\sin(\theta+k2\pi)\bigg)$ will be

$$\begin{align} % z_0 &= \sqrt[n]{r}\bigg(\cos(\theta/n) +i~\sin(\theta/n)\bigg) \\ % z_1 &= \sqrt[n]{r}\bigg(\cos(\frac{\theta}{n}+\frac{2\pi}{n}) +i~\sin(\frac{\theta}{n} +\frac{2\pi}{n}) \bigg) \\ % z_j &= \sqrt[n]{r}\bigg(\cos(\frac{\theta}{n}+\frac{2\pi j}{n}) +i~\sin(\frac{\theta}{n} +\frac{2\pi j}{n})\bigg) \text{ for all } j \leq n-1 \\ % \end{align}$$

So for your situation, $j=0,1$. Just do the computation and you will find the two answers in terms of $m$.

If someone finds a simpler method then please tell me.

DanielV
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