Question

Evaluate the following:
(i) $2x^3+2x^2-7x+72,\mathrm{when}x=\frac{3-5i}{2}$
(ii) $x^4-4x^3+4x^2+8x+44,\mathrm{when}x=3+2i$
(iii) $x^4+4x^3+6x^2+4x+9,\mathrm{when}x=-1+i\sqrt{2}$
(iv) $x^6+x^4+x^2+1,\mathrm{when}x=\frac{1+i}{\sqrt{2}}$
(v) $2x^4+5x^3+7x^2-x+41,\mathrm{when}x=-2-\sqrt{3}i$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)x=\frac{3-5i}{2} \\ \Rightarrow x^2={\left(\frac{3-5i}{2}\right)}^2 \\ =\frac{9+25i^2-30i}{4} \\ =\frac{-16-30i}{4} \\ \Rightarrow x^3=\frac{-16-30i}{4}\times \frac{3-5i}{2} \\ =\frac{-48+80i-90i+150i^2}{8} \\ =\frac{-198-10i}{8} \\ \therefore 2x^3+2x^2-7x+72=2\left(\frac{-198-10i}{8}\right)+2\left(\frac{-16-30i}{4}\right)-7\left(\frac{3-5i}{2}\right)+72 \\ =\frac{-198-10i-32-60i-42+70i+288}{4} \\ =\frac{16}{4} \\ =4 \\ \left(\mathrm{ii}\right)x=3+2i \\ \Rightarrow x^2={\left(3+2i\right)}^2 \\ =9+4i^2+12i \\ =5+12i \\ \Rightarrow x^3=x^2\times x \\ =\left(5+12i\right)\times \left(3+2i\right) \\ =15+10i+36i-24 \\ =-9+46i \\ \Rightarrow x^4={\left(x^2\right)}^2 \\ ={\left(5+12i\right)}^2 \\ =25+144i^2+120i \\ =-119+120i \\ \Rightarrow x^4-4x^3+4x^2+8x+44=-119+120i-4\left(-9+46i\right)+4\left(5+12i\right)+8\left(3+2i\right)+44 \\ =-119+120i+36-184i+20+48i+24+16i+44 \\ =5 \\ \left(\mathrm{iii}\right)x=-1+\sqrt{2}i \\ \Rightarrow x^2={\left(-1+\sqrt{2}i\right)}^2 \\ =1+2i^2-2\sqrt{2}i \\ =-1-2\sqrt{2}i \\ \Rightarrow x^3=\left(-1-2\sqrt{2}i\right)\times \left(-1+\sqrt{2}i\right) \\ =1-\sqrt{2}i+2\sqrt{2}i-4i^2 \\ =5+\sqrt{2}i \\ \Rightarrow x^4={\left(-1-2\sqrt{2}i\right)}^2 \\ =1+8i^2+4\sqrt{2}i \\ =-7+4\sqrt{2}i \\ \Rightarrow x^4+4x^3+6x^2+4x+9=-7+4\sqrt{2}i+4\left(5+\sqrt{2}i\right)+6\left(-1-2\sqrt{2}i\right)+4\left(-1+\sqrt{2}i\right)+9 \\ =-7+4\sqrt{2}i+20+4\sqrt{2}i-6-12\sqrt{2}i-4+4\sqrt{2}i+9 \\ =12 \\ \left(\mathrm{iv}\right)x=\frac{1+i}{\sqrt{2}} \\ \Rightarrow x^2={\left(\frac{1+i}{\sqrt{2}}\right)}^2 \\ =\left(\frac{1+i^2+2i}{2}\right) \\ =\frac{2i}{2} \\ =i \\ \Rightarrow x^6={\left(x^2\right)}^3 \\ =i^3 \\ =-i \\ \Rightarrow x^2=i \\ \Rightarrow x^4={\left(x^2\right)}^2 \\ =i^2 \\ =-1 \\ \mathrm{Now},x^6+x^4+x^2+1=-i-1+i+1 \\ =0 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)x=-2-\sqrt{3}i \\ \Rightarrow x^2={\left(-2-\sqrt{3}i\right)}^2 \\ =(-2{)}^2+(-\sqrt{3}i{)}^2+2(-2)(-\sqrt{3}i) \\ =4+3i^2+4\sqrt{3}i \\ =4-3+4\sqrt{3}i[\because i^2=-1] \\ =1+4\sqrt{3}i \\ \Rightarrow x^3=\left(1+4\sqrt{3}i\right)\times \left(-2-\sqrt{3}i\right) \\ =-2-\sqrt{3}i-8\sqrt{3}i-12i^2 \\ =10-9\sqrt{3}i[\because i^2=-1] \\ \Rightarrow x^4={\left(1+4\sqrt{3}i\right)}^2 \\ =1+48i^2+8\sqrt{3}i \\ =-47+8\sqrt{3}i[\because i^2=-1] \\ \Rightarrow 2x^4+5x^3+7x^2-x+41=2(-47+8\sqrt{3}i)+5\left(10-9\sqrt{3}i\right)+7\left(1+4\sqrt{3}i\right)-\left(-2-\sqrt{3}i\right)+41 \\ =-94+16\sqrt{3}i+50-45\sqrt{3}i+7+28\sqrt{3}i+2+\sqrt{3}i+41 \\ =6 \end{gathered}$$