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$az²+bz+10i=0$ where $a$ and $b$ are real, has a root of $3-i$. Show that $a=3$ and find the value of $b$.

I know that $3+i$ is also a root. If $a=3$, I found that $b$ would equal to $-12-2i$.

Please help. How come $a=3$?

  • 2
    `I know that 3+i is also a root` Why? – 2017-02-08
  • 6
    In general, *all* of the coefficients of your polynomial must be real in order to deduce that if $z \in \Bbb{C}$ is a root, then $\overline{z}$ is also a root. Here the constant coefficient is *not* real. – 2017-02-08
  • 0
    The other root is not $3+i,$ but the imaginary part of the other real root _is_ $i.$ Can you determine this from the given information about the root $3-i$ and about the coefficient $b$? – 2017-02-08
  • 2
    Didn't you notice, by the way, that $-12-2i$ is not real? – 2017-02-08

2 Answers 2

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Put z = 3 - i.

$a(3 - i)^2 + b(3 - i) + 10i = 0$

$a(8 - 6i) + b(3 - i) + 10i = 0$

$(8a + 3b) + (10 - 6a - b)i = 0$

Given a and b are real.

8a + 3b = 0 ........(1)

and 10 - 6a - b = 0 ......(2)

Substituting b = 10 - 6a in (1) gives,

8a + 3(10 - 6a) = 0.

8a + 30 - 18a = 0.

-10a + 30 = 0.

-10a = -30.

a = 3.

From (2),

10 - 6(3) - b = 0

10 - 18 - b = 0

-8 - b = 0.

b = -8.

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You can use the following facts:

If the two roots of $az^2+bz+c=0$ are $r_1$ and $r_2,$ then $$c = ar_1r_2 \tag1$$ and $$b = -a(r_1 + r_2).\tag2$$

Since $a$ and $b$ are real, from Equation $2$ we know $r_1+r_2$ is real. But since $c=10i$ is imaginary, from Equation $1$ we know $r_1r_2$ is imaginary.

Knowing $r_1 = 3-i$ and knowing that $r_1+r_2$ is real, we can get the imaginary part of $r_2.$ Knowing $r_1$ and knowing that $r_1r_2$ is imaginary, we can get the argument of $r_2.$ From those two facts we can find the exact value of $r_2$ (it is not $3+i$), and then from Equations $1$ and $2$ and the known values of $c,$ $r_1,$ and $r_2,$ we can find $a$ and then find $b.$