Question

Consider the equation $10z^2-3iz-k=0$, where z is a complex variable and $i^2=-1$. Which of the following statements is true?

• A. For all purely imaginary numbers $k$, both roots are real and irrational.
• B. For all complex numbers $k$, neither root is real.
• C. For real positive numbers $k$, both roots are purely imaginary.
• D. For real negative numbers $k$, both roots are purely imaginary.

Answer 1

The correct option is D

For real negative numbers $k$, both roots are purely imaginary.

Given: $10z^2-3iz-k=0$
It is in the form of $a{x}^2+bx+c=0$ whose roots are

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

$\Rightarrow z=\frac{-(-3i)\pm \sqrt{{(-3i)}^2-4\left(10\right)(-k)}}{2\left(10\right)}$

$\Rightarrow z=\frac{3i\pm \sqrt{-9+40k}}{20}$

Now, $D=-9+40k.Ifk=1,thenD=31.$
So, option (A) is false.
If k is a negative real number, then D is a negative real number.

So, option (D) is true.
If k = i, then

$D=-9+40i=16+40i-25={(4+5i)}^2,$

And the roots are $\left(\frac{1}{5}\right)+\left(\frac{2}{5}\right)iand-(\frac{1}{5}-\frac{1}{10})i.$

So, option (C) is false.
If k = 0 (which is a complex number), then the roots are 0 and

$\left(\frac{3}{10}\right)i.$
So, option (B) is false.