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Question: If \(x=1+i\) , then the value of the expression \({{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3\) is (a)-1...

If x=1+ix=1+i , then the value of the expression x4−4x3+7x2−6x+3{{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3 is
(a)-1
(b)1
(c)2
(d)None of these

Explanation

Solution

Hint: In this question, we can use the concept of power of Iota. Iota is square root of minus 1 means −1\sqrt{-1}. For example, what is the value of Iota's power 3? Iota i=−1i=\sqrt{-1} so i2{{i}^{2}} is −1×−1=−1\sqrt{-1}\times \sqrt{-1}=-1 and hence i3=i2×i=−i{{i}^{3}}={{i}^{2}}\times i=-i.

Complete step-by-step answer:
It is given that x=1+ix=1+i and let us consider the given expression, put x=1+ix=1+i in the given expression.
x4−4x3+7x2−6x+3=(1+i)4−4(1+i)3+7(1+i)2−6(1+i)+3.......(1){{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{{(1+i)}^{4}} by using (a+b)4=a4+4a3b+6a2b2+4ab3+b4{{\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+6i2+4i3+i4=1+4i−6−4i+1=−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{{(1+i)}^{3}} by using (a+b)3=a3+3a2b+3ab2+b3{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}
(1+i)3=1+3i+3i2+i3=1+3i−3−i=−2+2i{{(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{{(1+i)}^{2}} by using (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
(1+i)2=1+2i+i2=1+2i−1=2i{{(1+i)}^{2}}=1+2i+{{i}^{2}}=1+2i-1=2i
Now put all the values in the equation (1), we get
x4−4x3+7x2−6x+3=−4−4(−2+2i)+7(2i)−6(1+i)+3{{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
x4−4x3+7x2−6x+3=−4+8−8i+14i−6−6i+3{{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
x4−4x3+7x2−6x+3=(−4+8−6+3)+(−8i+14i−6i){{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=(-4+8-6+3)+(-8i+14i-6i)
Finally, we get
x4−4x3+7x2−6x+3=1+0i=1{{x}^{4}}-4{{x}^{3}}+7{{x}^{2}}-6x+3=1+0i=1
Hence the value of the given expression x4−4x3+7x2−6x+3{{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, 7i\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.