Question

Le a ∈R and f(x) = x³ 2x² + ax - a. Find the value of 'a' such that f(1 + i) is a real number. ​

Answer 1

Let,

$\longrightarrow x=1+i$

$\longrightarrow x-1=i$

Squaring both sides,

$\longrightarrow (x-1)^2=i^2$

$\longrightarrow x^2-2x+1=-1$

$\longrightarrow x^2-2x+2=0$

Now,

$\longrightarrow f(x)=x^3-2x^2+ax-a$

$\longrightarrow f(x)=x^3-2x^2+2x+ax-2x-a$

$\longrightarrow f(x)=x\left(x^2-2x+2\right)+(a-2)x-a$

For $x=1+i,$

$\longrightarrow f(1+i)=(1+i)(0)+(a-2)(1+i)-a$

$\longrightarrow f(1+i)=(a-2)i-2$

Given that $f(1+i)$ is a real number, so the coefficient of imaginary part here should be zero, i.e.,

$\longrightarrow a-2=0$

$\longrightarrow\underline{\underline{a=2}}$

Hence 2 is the answer.