2
$\begingroup$

Consider the equation with $z ∈ ℂ$ $$2z^2 − (3 + 8i)z − (m + 4i) = 0$$ where $m$ is a real constant and such that one of the two solutions is real.

I want to know how to calculate the solutions. I try to use $b^2-4ac$ to solve it, but it is hard to solve it. The unknown coefficient is so tricky

$\endgroup$
1
  • $\begingroup$ Welcome to MSE: Where did you get stuck? $\endgroup$ Commented Apr 21, 2018 at 7:54

4 Answers 4

1
$\begingroup$

If $r$ is a real solution, then $2r^2 − (3 + 8𝑖)r − (m + 4𝑖) = 0$. Hence, $2r^2-3r-m=0$ and $-8r-4=0$.

So, $\displaystyle r=-\frac{1}{2}$ and $m=2$.

The other root is $\displaystyle \frac{3+8i}{2}-\frac{-1}{2}=2+4i$.

$\endgroup$
2
  • $\begingroup$ I find r = -0.5. Am I wrong?? $\endgroup$
    – SteveYM
    Commented Apr 21, 2018 at 7:57
  • $\begingroup$ @SteveYM Yes. I made some mistakes. Thanks for pointing out. $\endgroup$
    – CY Aries
    Commented Apr 21, 2018 at 8:01
1
$\begingroup$

Hint. Let $x$ be the real solution and $z=a+ib$ be the other solution. Then the sum and the product of these solutions can be obtained from the coefficients of the quadratic: $$x+(a+ib)=\frac{3+8i}{2}\quad \mbox{and}\quad x(a+ib)=−\frac{m + 4𝑖}{2}.$$ Hence, since $m,x,a,b\in\mathbb{R}$, after separating real and imaginary parts, we get $$\begin{cases} x+a=\frac{3}{2}\\ b=4\\ xa=-\frac{m}{2}\\ xb=-2 \end{cases}$$ Can you take it from here?

$\endgroup$
1
  • $\begingroup$ Got it. Thank you $\endgroup$
    – SteveYM
    Commented Apr 21, 2018 at 7:58
0
$\begingroup$

Hint: by the quadratic formula we get $$z_{1,2}=\frac{1}{4}(3+8i)\pm\sqrt{\frac{1}{16}(3+8i)^2-\frac{1}{2}(m+4i)}$$ Can you finish? Second hint: we get the first solution as $$z_1=3/4+1/8\,\sqrt {2\,\sqrt { \left( -55+8\,m \right) ^{2}+6400}-110+16\, m}+i \left( 2+1/8\,\sqrt {2\,\sqrt { \left( -55+8\,m \right) ^{2}+6400 }+110-16\,m} \right) $$

$\endgroup$
0
0
$\begingroup$

HINT

We have

$$2z^2 − (3 + 8𝑖)z − (m + 4𝑖) = 0\iff z^2 − \frac{3 + 8𝑖}2z − \frac{m + 4𝑖}2 = 0$$

  • $z_1+z_2=\frac{3 + 8𝑖}2$

  • $z_1z_2=− \frac{m + 4𝑖}2$

let wlog $z_2=a\in \mathbb{R}$.

$\endgroup$
2
  • $\begingroup$ Thank you. I got the method $\endgroup$
    – SteveYM
    Commented Apr 21, 2018 at 7:59
  • $\begingroup$ You are welcome! Bye $\endgroup$
    – user
    Commented Apr 21, 2018 at 7:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .