Finding the roots of $x^2+(3+5i)x+(7+11i)=0$

Question

how can I solve following equation analytically

$$x^2+(3+5i)x+(7+11i)=0$$ I need the roots as follow $x=a+bi$

Answer 1

For quadratic equation in form of $$ax^2+bx+c=0$$ Using quadrtic formula we get

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Then for $$x^2+(3+5i)x+(7+11i)=0$$ we get $$x=\frac{-(3+5i)\pm\sqrt{(3+5i)^2-4(7+11i)}}{2}$$

$$x=\frac{-(3+5i)\pm\sqrt{-44-14i}}{2}$$

$$-44-14i=2\sqrt{533}\exp\left(\arctan\left(\frac{14}{44}\right)+\pi\right)$$

$$\sqrt{-44-14i}=\sqrt{2}{\sqrt[4]{533}\exp\left(\frac12\arctan\left(\frac{14}{44}\right)+\frac{\pi}{2}\right)}$$

$$x=\frac{-(3+5i)\pm\sqrt{2}{\sqrt[4]{533}\left[\cos\left(\frac12\arctan\left(\frac{14}{44}\right)+\frac{\pi}{2}\right)+i\sin\left(\frac12\arctan\left(\frac{14}{44}\right)+\frac{\pi}{2}\right)\right]}}{2}$$

I hope you can take it from here!

Answer 2

Consider the more general equation:

$$x^2+bx+c=0.$$

The quadratic formula gives:

$$x=\frac{-b\pm\sqrt{b^2-4c}}{2}.$$

Now substitute $b=3+5i$ and $c=7+11i$ and simplify. This gives

$$x=\frac{-(3+5i)\pm(-44-14i)^{1/2}}{2}.$$

We can write this as

$$x=\frac{-(3+5i)\pm e^{1/2\log(-44-14i)}}{2}.$$

But $$e^{1/2\log(-44-14i)} = e^{1/2\log\sqrt{44^2+14^2}+1/2i\text{Arg}(-44-14i)},$$

which is equal to

$$2132^{1/4}(\cos(-\pi/2+\tan^{-1}(7/22)/2)+i\sin(-\pi/2+\tan^{-1}(7/22)/2)).$$

This "simplifies" to

$$2132^{1/4}(\sin(\tan^{-1}(7/22)/2)-i\cos(\tan^{-1}(7/22)/2)).$$

Hence, your $x$ values are:

$$x=\frac{-(3+5i)\pm2132^{1/4}(\sin(\tan^{-1}(7/22)/2)-i\cos(\tan^{-1}(7/22)/2))}{2}.$$

Expanding gives:

$$x=-\frac{3}{2}+\frac{\sqrt[4]{533} \sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{7}{22}\right)\right)}{\sqrt{2}}+i \left(-\frac{5}{2}-\frac{\sqrt[4]{533} \cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{7}{22}\right)\right)}{\sqrt{2}}\right),$$

and

$$x=-\frac{3}{2}-\frac{\sqrt[4]{533} \sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{7}{22}\right)\right)}{\sqrt{2}}+i \left(\frac{\sqrt[4]{533} \cos \left(\frac{1}{2} \tan ^{-1}\left(\frac{7}{22}\right)\right)}{\sqrt{2}}-\frac{5}{2}\right).$$

Answer 3

An other solution to solve this is to put $x=a+ib$, $(a,b) \in \mathbb{R}$

Then $x²+(3+5i)x+(7+11i)=0$ becomes

$a²-b²+3a-5b+7+i(2ab+3b+5a+11)=0$

The real and imaginary parts must be each equal to $0$ and you compute like this.

A bit ugly, but a different way to do...

Answer 4

First of all,

$x=\frac{-3-5i\pm\sqrt{-44-14i}}{2}$

Let, $\sqrt{-44-14i}=(a+ib)$

Squaring both sides.

$-44-14i=a^2-b^2+2abi$

Compare real and imaginary parts to find $a$ and $b$

$a^2-b^2=-44$ $(i)$
$2ab=-14$

Using,

$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$
$(a^2+b^2)=2.\sqrt{533}$ -$(ii)$

Solving $(i)$ and $(ii)$

$a=(\sqrt{\sqrt{533}-22})$ and $b=(\sqrt{\sqrt{533}+22})$

Substituting value of $\sqrt{-44-14i}$ as $(a+ib)$ in the value of $x$.

I found the answer as,

$x=\frac{-3-5i\pm(\sqrt{\sqrt{533}-22} +\sqrt{\sqrt{533}+22}i )}{2}$

Answer 5

If you have to solve $$x^2+(3+5i)x+(7+11i)=0$$ and search for $x=a+ib$, just replace in the equation which, after developing and simplifications, leads to $$a^2+3 a-b^2-5 b+7+i (2 a b+5 a+3 b+11)=0$$ So, you have to set to $0$ the real and imaginary parts and you so have two equations for two unknowns $a$ and $b$ $$a^2+3 a-b^2-5 b+7=0$$ $$2 a b+5 a+3 b+11=0$$ which are quite ugly.

If you eliminate $b$ from the second equation, you have $$b=-\frac{5 a+11}{2 a+3}$$ Replacing in the first, reducing to same denominator leads to $$4 a^4+24 a^3+98 a^2+186 a+107=0$$ which is a quartic and then, by definition, very difficult to solve.

I am surprized you have been asked for the solution of such an equation.

Any way, the two real solutions are $$a_{\pm}=\frac{1}{2} \left(-3\pm\sqrt{\sqrt{533}-22}\right)$$ I let you the pleasure of finding the corresponding $b$'s (they are really bad !).

Answer 6

Start with

$$P \equiv \bigg(x^2 + (3+5i)x + (7 + 11i) = 0\bigg)$$

Substitute $x = a+bi$ into $P$, assuming $a$ and $b$ are $\mathbb R$, expand and equate the real and complex parts of the equation to get:

$$\begin{cases} P_1 \equiv \bigg(2ab + 3b + 5a + 11 = 0\bigg) \\ P_2 \equiv \bigg(-b^2 -5b + a^2 + 3a + 7 = 0\bigg) \end{cases}$$

Solve $P_1$ for $a$ and plug into $P_2$, you get a biquadratic in $b$, solve it.

Solve $P_1$ for $b$ and plug into $P_2$, you get a biquadratic in $a$, solve it.

$$a = \frac { \pm \sqrt{ 22 \pm \sqrt{533}}}2 - \frac 52$$ $$b = \frac { \pm \sqrt{ 22 \pm \sqrt{533}}}2 - \frac 32$$

Checking the results, the two solutions are:

$$x = \frac{ -\left|\sqrt{-22 - \left|\sqrt{533}\right|}\right| - 3 + i\bigg(-\left|\sqrt{ 22 - \left|\sqrt{533}\right|}\right| - 5\bigg) } 2$$

$$x = \frac{ \left|\sqrt{-22 - \left|\sqrt{533}\right|}\right| - 3 + i\bigg(-\left|\sqrt{ 22 - \left|\sqrt{533}\right|}\right| - 5\bigg) } 2$$

In hindsight (knowing the solution), the problem is much easier to solve if you make the substitution $x = (c - 3/2) + (d - 5/2)i$. Using that substitution you only need to solve quadratics.