Find the two complex numbers whose sum is $5 - i$ and product is $8+i$

Question

I have tried to solve this problem, but when I write the equations I end up with:

$$a + c = 5$$ $$b + d = -1$$ $$ab - cd = 8$$ $$ad + cb = 1$$

But substituting $a$ and $b$ get 2 equations I can't be able to solve

Answer 1

Hint: the numbers are the solutions to the equation $z^2 - (5-i)z + (8+i) =0$. (by Vieta's formulas)

Answer 2

We need to solve $$z^2-(5-i)z+8+i=0.$$ Now, $$\Delta=(5-i)^2-4(8+i)=24-10i-32-4i=-8-14i=(a+bi)^2,$$ where $a$ and $b$ are reals.

I got $$\Delta=\left(\sqrt{\sqrt{65}-4}-\sqrt{\sqrt{65}+4}i\right)^2.$$

Can you end it now?

Answer 3

$z+w=5-i$

$zw=8+i$

$w=\dfrac{8+i}z$

$z+\dfrac{8+i}z=5-i$

$z^2-(5-i)z+(8+i)=0$

Can you take it from here?

Answer 4

Get used to thinking of complex numbers as single entities and not as two numbers glued togeter. Do this with $2$, not $4$ variables.

$z + w = 5-i$ so $z = (5-i) - w$.

And $zw = (8+i)$ so $[(5-i)-w]w = (8+i)$ so

$-w^2 + (5-i)w - (8+i) = 0$

$w^2 - (5-i)w + (8+i) = 0$.

So just use the quadratic formula:

$w = \frac {-(5-i) \pm \sqrt {(5-i)^2 -4(8+i)}}{2}$

Now:

$(5-i)^2 = (25-1) -2*5i=24-10i$ and $4(8+i)=32+4i$ so $(5-i)^2 -4(8+i)=-8-14i$.

Okay finding the square roots of $-8-14i$ might be hard. It's easy if you know polar coordinates but I'll assume you haven't learned those yet.

If $(a+bi)^2 = -8 - 14i$ then $a^2 - b^2 = (a-b)(a+b)=-8$ and $2ab = -14$ so $ab=-7$.

This is rather unpleasant but.

If $a = -\frac 7b$ then $\frac {49}b^2 -b^2 + 8 = 0$ so $b^4-8b^2 -49 = 0$ and we can use the quadratic formula to solve for four possible value of $b$.