0
$\begingroup$

I stumbled upon the following exercise:

Find complex roots of a polynomial $x^2 - 5x + 7 - i$. The problem I have with this is not in the fact that I cannot solve it (I actually did), but I am searching for a more efficient or "smart" way of doing this.

The way I approached this problem was by writing x as a complex number ($x = a + bi$) and then substituted $x$ in the equation $x^2 - 5x + 7 - i = 0$. Then, after some rearranging I got two equations and solved the system for a and b. My result was correct, however, I am not satisfied because my method is rather tedious and I would like to hear from you, how to make this easier if possible.

Note: I have tried solving this as a quadratic equation the usual way, however, I stumbled upon $\sqrt{4i - 3}$ in the solution and I am not sure how to simplify this.

$\endgroup$
1
  • $\begingroup$ Stupid downvote. $\endgroup$
    – user65203
    May 15, 2018 at 9:40

4 Answers 4

0
$\begingroup$

Hint. Find $z=a+ib$ such that $z^2=\Delta=4i-3$: $$z^2=a^2-b^2+2iab=-3+4i\implies\begin{cases}a^2-b^2=-3\\ab=2\end{cases} \implies z=\pm(1+2i).$$ Now it should be easy to solve the quadratic equation $x^2 - 5x + 7 - i = 0$ in the "usual way".

$\endgroup$
0
$\begingroup$

I'll show you how to do $\sqrt{3-4i}$.

Let $(a+bi)^2 =c+di=z$

Then it can be derived that $$a=\pm\sqrt{\frac{c\pm|z|}{2}}, b=\pm\sqrt{\frac{-c\pm|z|}{2}}$$

For $z=-3+4i, |z|=\sqrt{3^2+4^2}=5$

Hence $$a=\pm\sqrt{\frac{3\pm5}{2}}=\pm i, \pm2$$ $$b=\pm\sqrt{\frac{-3\pm5}{2}}=\pm 1, \pm2i$$ Then you just need to play around with these (square them!) and find that $(1+2i)$ and $(-1-2i)$ are the square roots of $3-4i$

$\endgroup$
1
  • $\begingroup$ If you'd like me to explain how I got those formulae, just ask and I'll update my answer :) $\endgroup$ May 15, 2018 at 9:43
0
$\begingroup$

There is in fact no simpler way.

You can solve with the classical formulas and end up with the square root of a complex number. Then you can extract the square root using the polar form and the half-angle formulas*.

But this is not really simpler than the direct "plug $a+ib$" method that leads to a system of real quadratic equations in all cases.


*$$a+ib=\sqrt{a^2+b^2}(\cos\theta+i\sin\theta)$$ and

$$\sqrt{a+ib}=\sqrt[4]{a^2+b^2}\left(\cos\frac\theta2+i\sin\frac\theta2\right)\\ =\sqrt[4]{a^2+b^2}\left(\sqrt{\frac{1+\cos\theta}2}+i\sqrt{\frac{1-\cos\theta}2}\right)\\ =\sqrt[4]{a^2+b^2}\left(\sqrt{\frac{1+\dfrac a{\sqrt{a^2+b^2}}}2}+i\sqrt{\frac{1-\dfrac a{\sqrt{a^2+b^2}}}2}\right)\\ =\sqrt{\frac{\sqrt{a^2+b^2}+a}2}+i\sqrt{\frac{\sqrt{a^2+b^2}-a}2}.$$

$\endgroup$
0
$\begingroup$

Completing the square: $$x^2 - 5x + 7 - i=0 \iff \left(x-\frac52\right)^2=\frac{25}{4}-7+i \iff \\ x-\frac52=\pm\sqrt{-\frac34+i}=\pm (a+bi).$$ Note: $$(a+bi)^2=-\frac34+i \Rightarrow a^2-b^2+2abi=-\frac34+i \Rightarrow \\ \begin{cases}a^2-b^2=-\frac34\\ 2ab=1\end{cases}\Rightarrow (a,b)=\pm\left(\frac12,1\right).$$ Hence: $$x-\frac52=\pm\left(\pm\left(\frac12+i\right)\right)= \pm\left(\frac12+i\right)\Rightarrow\\ x_1=\frac52+\frac12+i=3+i;\\ x_2=\frac52-\frac12-i=2-i.$$

$\endgroup$

You must log in to answer this question.

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