Find the solution of $z^2−(3+i)z +4+3i = 0$

Question

Find the solution of the following equation: $$z^2−(3+i)z +4+3i = 0.$$

I calculated $\Delta = -6i-8$, and to find $\sqrt{\Delta}=\sqrt{-6i-8}$ we need the modulus and the argument of $-6i-8$. We have $|-6i-8|=10$, $\sin(\phi)=-6/10$, $\cos(\phi)=-8/10$, so $\phi=\arctan(3/4)$. Unfortunately, it makes all the computation quite complicated...

I can see on wolframalpha.com, that the solution is of a simpler form, https://www.wolframalpha.com/input?i=z+%5E2%E2%88%92%283%2Bi%29z+%2B4%2B3i+%3D+0, but I can not see how I can simplify the equation to get such a simple form. I would be grateful for your hints.

Answer 1

I don't see how the computation is "quite complicated." Here's a general method to approach such problems (remembering only one double angle formula).

By the double angle formula, $\cos\phi = 2\cos^2(\phi/2) - 1$, so $2\cos^2(\phi/2) = 1/5$ and $\cos(\phi/2) = \pm 1/\sqrt{10}$. Thus $\text{Re}(\sqrt\Delta) = \sqrt{10}(\pm 1/\sqrt{10}) = \pm 1$. Continuing, $\sin(\phi/2) = \pm 3/\sqrt{10}$, and so $\text{Im}(\sqrt\Delta) = \pm 3$. Since the values of $\sqrt\Delta$ lie in quadrants II and IV, we have $$\sqrt\Delta = \pm(-1+3i).$$

Answer 2

A different way is to show that $$ z^2 - (3 + i) \, z + (4 + 3 \, i) = 0$$ can be factored into the form $$ (z + a + b \, i)(z + c + d \, i) = 0 $$ and takes the form $$ \left(z - \frac{3 \, (1+b)}{1+2 b} + b \, i\right)\left(z - \frac{3 \, b}{1+2 b} - (1+b) \, i\right) = 0.$$ This gives the two roots as $$ z = \frac{3 \, (1+b)}{1+2 b} - b \, i \hspace{5mm} \text{and} \hspace{5mm} z = \frac{3 \, b}{1+2 b} + (1+b) \, i. $$ By comparing the equations it is seen that $b = 1$ and the roots are quickly obtained.

Answer 3

$\sin\theta = -0.6\\ \cos\theta = -0.8$

$\tan \frac \theta 2 = \frac {\sin \theta}{1+\cos \theta} = \frac {-0.6}{0.2} = -3$

$\sqrt {-8-6i} = a(1-3i)$

To find $a$
$|-8-6i| = 10\\ |1-3i| = \sqrt {10}\\ a = 1$

Answer 4

$(z-(1+2i))(z-(2-i))=0$

I found this by completing square (Best way ofcourse): $(z-\frac{3+i}{2})^2=\pm\frac{i}{\sqrt{2}}\sqrt{4+3i}=\pm\frac{i}{\sqrt{2}}\frac{3+i}{\sqrt{2}}$ and thus $(z-\frac{3+i}{2})^2=\pm\frac{-1+3i}{2}$ hence the rest is easier...

... I MIXED HERE. Shame on me. :( Disscriminant of the quadratic equation: $$\Delta=(-(3+i))^2-4(1)(4+3i)=9+6i-1-16-12i=-8-6i=re^{i\phi}$$ where $r=\sqrt{(-8)^2+(-6)^2}=10$ and $\phi=\pi+\arctan(\frac{3}{4})$, because it is in 3-rd region. So, the roots are $$z_{1,2}=\frac{3+i\pm\sqrt{10e^{i\phi}}}{2}=\frac{3+i\pm\sqrt{10}e^{i\frac{\phi}{2}}}{2}= \frac{3+i\pm\sqrt{10}ie^{i\frac{\arctan(3/4)}{2}}}{2} = \frac{3+i\pm\sqrt{10}i\left(\frac{3}{\sqrt{10}}+i\frac{1}{\sqrt{10}})\right)}{2} =\frac{3+i\pm i\left(3+i)\right)}{2}=\frac{3+i\pm(-1+3i)}{2}.$$ Hence roots are $1+2i$ and $2-i$.