Question

The roots of the equation $z^4+a z^3+(12+9 i) z^2+b z=0$ (where $a$ and $b$ are complex numbers) are the vertices of a square, then find the value of $|2 b-13 a|$.

Answer 1

Note that $z=0$ is one root of the given equation. $z^4+a z^3+(12+9 j) z^2+b z=0$
cancelling $z$, we get
$z ^3+ az ^2+(12+9 i) z + b =0$....(1)
The 3 vertices of the square other than $(0+0 i)$ are $u , i u , u (1+i)$ (think!) Now, $u ( u i)+(i u )(1+i) u +[(1+i) u ] u =12+9 i$ (using theory of equation) $u ^2[i+i-1+1+i]=12+9 i$
$3 i u ^2=12+9 i$
$u ^2=\frac{4+3 i}{i}=3-4 i \Rightarrow u =\sqrt{3-4 i }= \pm(2- i )$
$\therefore u =2-i$ or $-2+i$ (both values of $u$ will give the same result)
$\text { again, }- a = u + u i+ u (1+i)= u [1+i+1+i]=2 u (1+i)=2(2-i)(1+i)=2(3+i)=6+2 i$
$a =-6-2 i$
Again $( u )( u i)(1+i) u =- b$
$- b = u ^3(-1+i) $
$b =(2-i)^3(1-i)=\left[8-i^3-6 i(2-i)\right](1-i)=(8+i-12 i-6)(1-i)=(2-11 i)(1-i)$
$b =2-2 i-11 i-11=-9-13 i $
$\therefore b =-9-13 i$
Hence, $2 b -13 a =2(-9-13 i)-13(-6-2 i)=60$.
Note: If we take $u=-(2-i)$ then $a=6+2 i ; b=9+13 i$.
Then $(2 b-13 a)=-60$ Ans. $]$
[18-05-2014, CC JEE Adv, P-2]