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Question: \[\text{If }z=2-i\sqrt{7},\text{ then show that }3{{z}^{3}}-4{{z}^{2}}+z+88=0\]...

If z=2−i7, then show that 3z3−4z2+z+88=0\text{If }z=2-i\sqrt{7},\text{ then show that }3{{z}^{3}}-4{{z}^{2}}+z+88=0

Explanation

Solution

Hint: First try to solve the complex number by eliminating i from it, then get a quadratic equation in z without the i term. Now, try to get the cubic equation from the quadratic equation you found. Now, substitute this quadratic equation as 0. Then you also get the cubic equation as 0. Thus, we get the required proof.

Complete step-by-step answer:
Let us consider the given complex number in the question.
z=2−i7.....(i)z=2-i\sqrt{7}.....\left( i \right)
We need to prove the given cubic equation which is written as
3z3−4z2+z+88=0.....(ii)3{{z}^{3}}-4{{z}^{2}}+z+88=0.....\left( ii \right)
By subtracting 2 from both the sides of the equation (i), we get,
z−2=2−i7−2z-2=2-i\sqrt{7}-2
By canceling the common terms, we can write the equation as
z−2=−i7z-2=-i\sqrt{7}
By squaring both the sides of the equation, we get it as,
(z−2)2=(−i7)2{{\left( z-2 \right)}^{2}}={{\left( -i\sqrt{7} \right)}^{2}}
By basic properties of the power, we get to know the relation:
(ab)n=an.bn{{\left( ab \right)}^{n}}={{a}^{n}}.{{b}^{n}}
By applying this to the right-hand side, we get it as,
(z−2)2=i2(7)2{{\left( z-2 \right)}^{2}}={{i}^{2}}{{\left( \sqrt{7} \right)}^{2}}
By using (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} on the left-hand side and substituting i2=−1{{i}^{2}}=-1 on the right-hand side, we get,
z2−4z+4=−7{{z}^{2}}-4z+4=-7
By adding 7 on both the sides, we get the equation in the form of,
z2−4z+11=0....(iii){{z}^{2}}-4z+11=0....\left( iii \right)
By multiplying 3z on both the sides of the equation, we get,
3z(z2−4z+11)=03z\left( {{z}^{2}}-4z+11 \right)=0
By adding 88 and subtracting 88, we get the equation in the form,
3z3−12z2+33z+88−88=03{{z}^{3}}-12{{z}^{2}}+33z+88-88=0
Now, write −12z2 as −8z2−4z2-12{{z}^{2}}\text{ as }-8{{z}^{2}}-4{{z}^{2}} to get the cubic equation. We get,
3z3−4z2−8z2+33z+88−88=03{{z}^{3}}-4{{z}^{2}}-8{{z}^{2}}+33z+88-88=0
Now write 33z as z + 32z to get the cubic equation. We get,
3z3−4z2−8z2+z+32z+88−88=03{{z}^{3}}-4{{z}^{2}}-8{{z}^{2}}+z+32z+88-88=0
Now, taking the cubic equation together, we get the equation as,
(3z3−4z2+z+88)−8z2+32z−88=0\left( 3{{z}^{3}}-4{{z}^{2}}+z+88 \right)-8{{z}^{2}}+32z-88=0
By taking – 8 common from the last three terms, we get it as,
(3z3−4z2+z+88)−8(z2−4z+11)=0\left( 3{{z}^{3}}-4{{z}^{2}}+z+88 \right)-8\left( {{z}^{2}}-4z+11 \right)=0
By substituting the value of the quadratic equation from equation (iii), we get,
(3z3−4z2+z+88)−8(0)=0\left( 3{{z}^{3}}-4{{z}^{2}}+z+88 \right)-8\left( 0 \right)=0
By simplifying, we get the value of the cubic equation as
3z3−4z2+z+88=03{{z}^{3}}-4{{z}^{2}}+z+88=0
Hence proved.

Note: The idea is to multiply the terms, add, subtract 88 are derived to generate the given cube. This is not done by any trick but by observation only. While removing the brackets, students forget to multiply with the constant term. So, be careful at these steps as the whole answer depends on them. Also, while taking – 8 common from the last three terms, you will get + 11, in confusion students write it as – 11 and get a constant instead of 0.