Question

If $f\left(x\right)=4x^5+5x^4-8x^3+5x^2+4x-34i$ and $f\left(\frac{-1+\sqrt{3}i}{2}\right)$ equals to $a+ib$, then $a:b$ equal

• A. $1:2$
• B. $-2:1$
• C. $17:1$
• D. $-17:1$

Answer 1

The correct option is A

$1:2$

Explanation for the correct option.

Step 1. Find the function value $f\left(\frac{-1+\sqrt{3}i}{2}\right)$.

Here, $\frac{-1+\sqrt{3}i}{2}$ is equal to $\omega$ such that ${\omega }^3=1$ and $\omega +{\omega }^2=-1$.

So, for the function $f\left(x\right)=4x^5+5x^4-8x^3+5x^2+4x-34i$, the function value $f\left(\frac{-1+\sqrt{3}i}{2}\right)$ is given as:

$\begin{array}{rcl}f\left(\frac{-1+\sqrt{3}i}{2}\right)& =& f\left(\omega \right)\\ & =& 4{\omega }^5+5{\omega }^4-8{\omega }^3+5{\omega }^2+4\omega -34i\\ & =& 4{\omega }^3{\omega }^2+5{\omega }^3\omega -8{\omega }^3+5{\omega }^2+4\omega -34i\\ & =& 4{\omega }^2+5\omega -8+5{\omega }^2+4\omega -34i\mathbf{}\left[{\omega }^3=1\right]\\ & =& 9{\omega }^2+9\omega -8-34i\\ & =& 9\left({\omega }^2+\omega \right)-8-34i\\ & =& 9\left(-1\right)-8-34i\mathbf{}\mathbf{}\left[{\omega }^2+\omega =-1\right]\\ & =& -9-8-34i\\ & =& -17-34i\end{array}$

Step 2. Find the value of $a:b$.

On comparing $-17-34i$ with $a+ib$, it is found that $a=-17$ and $b=-34$. Now,

$\begin{array}{rcl}\frac{a}{b}& =& \frac{-17}{-34}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct option is A.