Is my result to $z^2 - (3+4i)z - 1 + 7i = 0$ right?

Question

I solved it much like I would a second degree equation, with a=1, b=-(3+4i) and c=(-1+7i). Is that the correct approach?

My final result was $z = \frac{3+4i \pm i\sqrt{3+4i}}{2}$

Is that correct? There is no way for me to check that.

Answer 1

Solving it using the quadratic formula is the way to go:

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

where your equation is $az^2 + bz + c = 0.$

You can always check your answer by substituting the solution back into the equation. But your value of $b$ is actually $-(3+4i)$ so your answers aren't correct if you implemented the quadratic formula correctly. (It may just as easily be a typo in your question.)

Answer 2

You are not done yet. There is no reason to leave complex numbers under the radical.

$z = \frac {(3+4i) \pm i\sqrt {3 + 4i}}2$

$3 + 4i = 5 e^{i\theta}\\ \sqrt {+3 +4i} = \sqrt 5 e^{i\frac {\theta}{2}}$

$e^{i\theta} = \cos\theta + i \sin\theta\\ e^{i\frac{\theta}2} = \cos\frac{\theta}2 + i \sin\frac{\theta}2\\ $

$\cos\theta = \frac 35\\ \cos\frac{\theta}2 = \sqrt {\frac {1+\frac 35}{2}} = \frac {2}{\sqrt 5}\\ \sin\frac{\theta}2 =-\sqrt {\frac {1-\frac 35}{2}} = -\frac 1{\sqrt 5}$

$\sqrt {3 +4i} = 2 +i$

$z = \frac {(3+4i) \pm i(2+i)}2$

$z = 2+i,1+3i$

Answer 3

Here is the standard way t o compute the square roots of a complex number: $$(x+iy)^2=3+4i\iff \begin{cases}x^2-y^2=3,\\2xy=4.\end{cases}$$ Further the square of the module of $x+iy$ is equal to the module of $3+4i$: $$x^2+y^2=5.$$ So we first solve the linear system in $x^2$ and $y^2$: $$\begin{cases}x^2-y^2=3,\\x^2+y^2=5\end{cases}\iff\begin{cases}x^2=4,\\y^2=1\end{cases}\iff\begin{cases}x=\pm2,\\y^2=\pm1.\end{cases}$$ Further,$xy=2>0$ implies $x$ and $y$ have the same sign, and finally we obtain $$x+iy=\pm(2+i).$$

Answer 4

Checking the roots is in principle just a matter of arithmetic. But it would be very easy to make a mistake with all those $i$'s and square roots.

An easier computation is to check that the sum and product of the two roots is correct. Given the equation $z^2+bx+c=0$, the sum of the roots is $-b$ and the product of the roots is $c$.

In this case, the sum of your two proposed roots is clearly $3+4i$, so that checks out.

So now you have to check the product: i.e. you have to check that $$\frac14\{(3+4i)^2-(i\sqrt{3+4i})^2\} = -1+7i$$ Evaluating the two squared terms: $$\frac14\{(-7+24i)+(3+4i)\} = -1+7i$$ By my reckoning, this equation is correct. So those are indeed the correct roots.