0
$\begingroup$

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?

$\endgroup$
17
  • $\begingroup$ Where is the sign of equality? $\endgroup$
    – Olimjon
    Jan 6, 2016 at 6:08
  • $\begingroup$ @Olimjon Oh sorry it's equals to 0 $\endgroup$
    – Amine Marzouki
    Jan 6, 2016 at 6:11
  • $\begingroup$ 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 $\mathbb{C}$. $\endgroup$
    – Jendrik Stelzner
    Jan 6, 2016 at 6:16
  • 1
    $\begingroup$ $$c = (m^2 + 1)i = (m+i)(m-i)i = (-1 + mi)(m - i) = (m + i)(1+mi)$$ $$b = (1 + m)(1 + i) = 1 + m + (m + 1)i$$ since you want $z_1z_2 = c$ and $z_1 + z_2 = b$ then you can maybe guess that $z_1 = m+i$ and $z_2 = 1 + mi$. $\endgroup$
    – DanielV
    Jan 6, 2016 at 6:48
  • 1
    $\begingroup$ @AmineMarzouki Whether that's it depends on what method you are trying to learn. $\endgroup$
    – DanielV
    Jan 6, 2016 at 6:52

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I'll tell you later :D we solved it Thanks for your answer too ! $\endgroup$
    – Amine Marzouki
    Jan 17, 2016 at 11:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .