
If $x=1+i$ , then the value of the expression ${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3$ is
(a)-1
(b)1
(c)2
(d)None of these
Answer
594.6k+ views
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.
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3={{(1+i)}^{4}}-4{{(1+i)}^{3}}+7{{(1+i)}^{2}}-6(1+i)+3.......(1)$
First work out on the expansion ${{(1+i)}^{4}}$ by using ${{\left( a+b \right)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}$
${{(1+i)}^{4}}=1+4i+6{{i}^{2}}+4{{i}^{3}}+{{i}^{4}}=1+4i-6-4i+1=-4$
Second work out on the expansion ${{(1+i)}^{3}}$ by using ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
${{(1+i)}^{3}}=1+3i+3{{i}^{2}}+{{i}^{3}}=1+3i-3-i=-2+2i$
Third work out on the expansion ${{(1+i)}^{2}}$ by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
${{(1+i)}^{2}}=1+2i+{{i}^{2}}=1+2i-1=2i$
Now put all the values in the equation (1), we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=-4-4(-2+2i)+7(2i)-6(1+i)+3$
Rearranging the terms, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=-4+8-8i+14i-6-6i+3$
Collecting all constant terms and iota terms, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=(-4+8-6+3)+(-8i+14i-6i)$
Finally, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=1+0i=1$
Hence the value of the given expression ${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3$ 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.
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.
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3={{(1+i)}^{4}}-4{{(1+i)}^{3}}+7{{(1+i)}^{2}}-6(1+i)+3.......(1)$
First work out on the expansion ${{(1+i)}^{4}}$ by using ${{\left( a+b \right)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}$
${{(1+i)}^{4}}=1+4i+6{{i}^{2}}+4{{i}^{3}}+{{i}^{4}}=1+4i-6-4i+1=-4$
Second work out on the expansion ${{(1+i)}^{3}}$ by using ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
${{(1+i)}^{3}}=1+3i+3{{i}^{2}}+{{i}^{3}}=1+3i-3-i=-2+2i$
Third work out on the expansion ${{(1+i)}^{2}}$ by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
${{(1+i)}^{2}}=1+2i+{{i}^{2}}=1+2i-1=2i$
Now put all the values in the equation (1), we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=-4-4(-2+2i)+7(2i)-6(1+i)+3$
Rearranging the terms, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=-4+8-8i+14i-6-6i+3$
Collecting all constant terms and iota terms, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=(-4+8-6+3)+(-8i+14i-6i)$
Finally, we get
${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=1+0i=1$
Hence the value of the given expression ${{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3$ 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.
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