Question

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

Answer 1

Answer: 0060

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