Question

If $x=1+i$ , then the value of the expression $x^4-4x^3+7x^2-6x+3$ is
(a)-1
(b)1
(c)2
(d)None of these

Answer 1

Hint: In this question, we can use the concept of power of Iota. Iota is square root of minus 1 means $\sqrt{-1}$. For example, what is the value of Iota's power 3? Iota $i=\sqrt{-1}$ so $i^2$ is $\sqrt{-1}\times \sqrt{-1}=-1$ and hence $i^3=i^2\times i=-i$.

Complete step-by-step answer:
It is given that $x=1+i$ and let us consider the given expression, put $x=1+i$ in the given expression.
$x^4-4x^3+7x^2-6x+3={(1+i)}^4-4{(1+i)}^3+7{(1+i)}^2-6(1+i)+3.......(1)$
First work out on the expansion ${(1+i)}^4$ by using ${(a+b)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$
${(1+i)}^4=1+4i+6i^2+4i^3+i^4=1+4i-6-4i+1=-4$
Second work out on the expansion ${(1+i)}^3$ by using ${(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3$
${(1+i)}^3=1+3i+3i^2+i^3=1+3i-3-i=-2+2i$
Third work out on the expansion ${(1+i)}^2$ by using ${(a+b)}^2=a^2+2ab+b^2$
${(1+i)}^2=1+2i+i^2=1+2i-1=2i$
Now put all the values in the equation (1), we get
$x^4-4x^3+7x^2-6x+3=-4-4(-2+2i)+7(2i)-6(1+i)+3$
Rearranging the terms, we get
$x^4-4x^3+7x^2-6x+3=-4+8-8i+14i-6-6i+3$
Collecting all constant terms and iota terms, we get
$x^4-4x^3+7x^2-6x+3=(-4+8-6+3)+(-8i+14i-6i)$
Finally, we get
$x^4-4x^3+7x^2-6x+3=1+0i=1$
Hence the value of the given expression $x^4-4x^3+7x^2-6x+3$ is 1 and which is purely real number.
Therefore, the correct option for the given question is option (b).

Note: We might get confused by the word purely imaginary. A complex number is said to be purely imaginary if it has no real part. For example, 3i, i, 5i, $\sqrt{7}i$ and −12i are all examples of pure imaginary numbers, or numbers of the form bi, where b is a nonzero real number.