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Hello everyone I am having some difficulties understanding some concepts. I am trying to solve for the roots of a complex polynomial.

$$f(x)=(3i+1)x^{2}+(-6i-2)x+12$$

I tried to use the quadratic formula that is,

$$x= \frac{-(-2-6i) \pm \sqrt{(-6i-2)^{2}-4(3i+1)(12)}}{2(3+i)}$$

$$x= \frac{2+6i \pm \sqrt{-176-24i}}{6+2i}$$

Now is the point where I am confused, $$\sqrt{-176-24i}$$ which I rewrote as $$\sqrt{((-1)(8)(22+3i)}$$

and using what I thought were valid rules, wrote that as $$\sqrt{8}i\sqrt{22+3i}=2\sqrt{2}i\sqrt{22+3i}$$

so with that form I thought I could write,

$$x= \frac{2(1+3i \pm \sqrt{2}i\sqrt{22+3i})}{2(3+i)}$$

$$x= \frac{(1+3i \pm \sqrt{2}i\sqrt{22+3i})}{(3+i)}$$

But now I am stuck and am not sure if the form is correct, and I am not sure how I can proceed to get a final answer. Any help please? I am still stuck on this. Is there a way I can do it without changing into polar form? Also I appreciate all the hints for finding the square roots and such, but I am still not even sure my work is correct. Could someone help with that? It seems to disagree with wolfram.

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    $\begingroup$ i have got this here $$1-\sqrt{-\frac{1}{5}+\frac{18 i}{5}}$$ $\endgroup$ Sep 17 '15 at 17:34
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    $\begingroup$ and this here $$1+\sqrt{-\frac{1}{5}+\frac{18 i}{5}}$$ $\endgroup$ Sep 17 '15 at 17:34
  • $\begingroup$ you need to rewrite $-176-24i$ into polar coordinates and then you need to apply the complex root. $\endgroup$
    – MrYouMath
    Sep 17 '15 at 17:37
  • $\begingroup$ I agree with @Dr.SonnhardGraubner; part of the mystery is how the OP got -176-24i rather than -80-120i. $\endgroup$ Sep 23 '15 at 4:28
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In general, let $z=x+iy$ where $x$ and $y$ are real-valued numbers. We can express $z$ in Polar Coordinates as $z=\sqrt{x^2+y^2}e^{i\arctan(x,y)}$ where the Arctangent Function $\arctan(x,y)$ is given by

$$ \arctan(x,y)=\begin{cases} \arctan(y/x)&,x>0\\\\ \pi+\arctan(y/x)&,x<0,y>0\\\\ -\pi+\arctan(y/x)&,x<0,y<0\\\\ \pi/2&,x=0,y>0\\\\ -\pi/2&,x=0,y<0 \end{cases}$$

Then, the square root of $z$, $z^{1/2}$ is one of the two values

$$z^{1/2}=\pm (x^2+y^2)^{1/4}e^{i\arctan(x,y)/2} \tag 1$$

We can convert $z^{1/2}$ back to rectangular coordinates using Euler's Identity on $(1)$. Proceeding accordingly gives

$$\begin{align} z^{1/2}&=\pm (x^2+y^2)^{1/4}\left(\cos\left(\frac12 \arctan(x,y)\right)+i\sin\left(\frac12\arctan(x,y)\right)\right)\\\\ &=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}\pm i\,\text{sgn}(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\tag 2 \end{align}$$

where in arriving at $(2)$ we used the Half-Angle Formulae for the sine and cosine funcitons.


Applying $(2)$ to the specific problem at hand for which $z=22+3i$, we have

$$\begin{align} z^{1/2}&=\pm\left(\sqrt{\frac{\sqrt{(22)^2+(3)^2}+22}{2}}+i\text{sgn}(3)\sqrt{\frac{\sqrt{(22)^2+(3)^2}-22}{2}}\right)\\\\ &\approx 4.701255327631890 +i0.319063717127566 \end{align}$$

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We're looking for $a+ib$ such that $$(a+ib)^2=a^2-b^2+2abi=-176-24i.$$ That is $a^2-b^2=-176$ and $2ab=-24$. So $b=-12/a$ and $$a^4+176a^2-144=0,$$ which gives $$a^2=-88\pm\sqrt{7888}$$ hence $a=\pm 2\sqrt{-22+\sqrt{493}}.$ Now you can easily find $b$.

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  • $\begingroup$ How do you know that a is not zero when you divide to put in terms of b? And how did you get the 7888? $\endgroup$
    – Quality
    Sep 19 '15 at 0:32
  • $\begingroup$ Since $2ab=-24$ both $a$ and $b$ are nonzero. I got the $7888$ by using the formula to solve the quadratic equation $$x^2+17x-144=0.$$ $\endgroup$
    – Sonner
    Sep 19 '15 at 7:18
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Another nice formula for the complex square root of some $z\in \Bbb C$ is $$ \sqrt{z}=\pm\sqrt{|z|}·\frac{z+|z|}{\bigl|z+|z|\bigr|} $$ encoding the facts that the root is on the angle bisector of $z$ and the real axis and that the absolute value is the root of the absolute value of $z$.

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