Evaluate the following:
Question:

Evaluate the following:

(i) $2 x^{3}+2 x^{2}-7 x+72$, when $x=\frac{3-5 i}{2}$

(ii) $x^{4}-4 x^{3}+4 x^{2}+8 x+44$, when $x=3+2 i$

 

(iii) $x^{4}+4 x^{3}+6 x^{2}+4 x+9$, when $x=-1+i \sqrt{2}$

(iv) $x^{6}+x^{4}+x^{2}+1$, when $x=\frac{1+i}{\sqrt{2}}$

(v) $2 x^{4}+5 x^{3}+7 x^{2}-x+41$, when $x=-2-\sqrt{3} i$

Solution:

(i) $x=\frac{3-5 i}{2}$

$\Rightarrow x^{2}=\left(\frac{3-5 i}{2}\right)^{2}$

$=\frac{9+25 i^{2}-30 i}{4}$

$=\frac{-16-30 i}{4}$

$\Rightarrow x^{3}=\frac{-16-30 i}{4} \times \frac{3-5 i}{2}$

$=\frac{-48+80 i-90 i+150 i^{2}}{8}$

$=\frac{-198-10 i}{8}$

$\therefore 2 x^{3}+2 x^{2}-7 x+72=2\left(\frac{-198-10 i}{8}\right)+2\left(\frac{-16-30 i}{4}\right)-7\left(\frac{3-5 i}{2}\right)+72$

$=\frac{-198-10 i-32-60 i-42+70 i+288}{4}$

$=\frac{16}{4}$

= 4

(ii) $x=3+2 i$

$\Rightarrow x^{2}=(3+2 i)^{2}$

$=9+4 i^{2}+12 i$

$=5+12 i$

$\Rightarrow x^{3}=x^{2} \times x$

$=(5+12 i) \times(3+2 i)$

$=15+10 i+36 i-24$

$=-9+46 i$

$\Rightarrow x^{4}=\left(x^{2}\right)^{2}$

$=(5+12 i)^{2}$

$=25+144 i^{2}+120 i$

$=-119+120 i$

$\Rightarrow x^{4}-4 x^{3}+4 x^{2}+8 x+44=-119+120 i-4(-9+46 i)+4(5+12 i)+8(3+2 i)+44$

$=-119+120 i+36-184 i+20+48 i+24+16 i+44$

= 5

(iii) $x=-1+\sqrt{2} i$

$\Rightarrow x^{2}=(-1+\sqrt{2} i)^{2}$

$=1+2 i^{2}-2 \sqrt{2} i$

$=-1-2 \sqrt{2} i$

$\Rightarrow x^{3}=(-1-2 \sqrt{2} i) \times(-1+\sqrt{2} i)$

$=1-\sqrt{2} i+2 \sqrt{2} i-4 i^{2}$

$=5+\sqrt{2} i$

$\Rightarrow x^{4}=(-1-2 \sqrt{2} i)^{2}$

$=1+8 i^{2}+4 \sqrt{2} i$

$=-7+4 \sqrt{2} i$

$\Rightarrow x^{4}+4 x^{3}+6 x^{2}+4 x+9=-7+4 \sqrt{2} i+4(5+\sqrt{2} i)+6(-1-2 \sqrt{2} i)+4(-1+\sqrt{2} i)+9$

$=-7+4 \sqrt{2} i+20+4 \sqrt{2} i-6-12 \sqrt{2} i-4+4 \sqrt{2} i+9$

= 12

(iv) $x=\frac{1+i}{\sqrt{2}}$

$\Rightarrow x^{2}=\left(\frac{1+i}{\sqrt{2}}\right)^{2}$

$=\left(\frac{1+i^{2}+2 i}{2}\right)$

$=\frac{2 i}{2}$

$=i$

$\Rightarrow x^{6}=\left(x^{2}\right)^{3}$

$=i^{3}$

$=-i$

$\Rightarrow x^{2}=i$

$\Rightarrow x^{4}=\left(x^{2}\right)^{2}$

$=i^{2}$

$=-1$

Now, $x^{6}+x^{4}+x^{2}+1=-i-1+i+1$

= 0

$(\mathrm{v}) x=-2-\sqrt{3} i$

$\Rightarrow x^{2}=(-2-\sqrt{3} i)^{2}$

$=(-2)^{2}+(-\sqrt{3} i)^{2}+2(-2)(-\sqrt{3} i)$

$=4+3 i^{2}+4 \sqrt{3} i$

$=4-3+4 \sqrt{3} i$     $\left[\because i^{2}=-1\right]$

$=1+4 \sqrt{3} i$

$\Rightarrow x^{3}=(1+4 \sqrt{3} i) \times(-2-\sqrt{3} i)$

$=-2-\sqrt{3} i-8 \sqrt{3} i-12 i^{2}$

$=10-9 \sqrt{3} i$   $\left[\because i^{2}=-1\right]$

$\Rightarrow x^{4}=(1+4 \sqrt{3} i)^{2}$

$=1+48 i^{2}+8 \sqrt{3} i$

$=-47+8 \sqrt{3} i$   $\left[\because i^{2}=-1\right]$

$\Rightarrow 2 x^{4}+5 x^{3}+7 x^{2}-x+41=2(-47+8 \sqrt{3} i)+5(10-9 \sqrt{3} i)+7(1+4 \sqrt{3} i)-(-2-\sqrt{3} i)+41$

$=-94+16 \sqrt{3} i+50-45 \sqrt{3} i+7+28 \sqrt{3} i+2+\sqrt{3} i+41$

= 6

 

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