You visited us 1 times! Enjoying our articles? Unlock Full Access!

nth Root of a Complex Number

Evaluate the ...

Question

Evaluate the following :

$(i) 2 x^{3} + 2 x^{2} - 7 x + 72, w h e n x = \frac{3 - 5 i}{2} (i i) x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, w h e n x = 3 + 2 i (i i i) x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, w h e n x = - 1 + i \sqrt{2} (i v) x^{6} + x^{4} + x^{2} + 1, w h e n x = \frac{1 + i}{\sqrt{2}} (v) 2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, w h e n x = - 2 - \sqrt{3} i$

Open in App

Solution

$(i) We have, x = \frac{3 - 5 i}{2} \Rightarrow 2 x = 3 - 5 i \Rightarrow 2 x - 3 = - 5 i \Rightarrow (2 x - 3)^{2} = (- 5 i)^{2} \Rightarrow 4 x^{2} + 9 - 12 x = - 25 \Rightarrow 4 x^{2} - 12 x + 34 = 0 \Rightarrow 2 (2 x^{2} - 6 x + 17) = 0 \Rightarrow 2 x^{2} - 6 x + 17 = 0 . . . (i) ∴ 2 x^{3} + 2 x^{2} - 7 x + 72 = x (2 x^{2} - 6 x + 17) + 6 x^{2} - 17 x + 2 x^{2} - 7 x + 72 (adding and subtracting ) 6 x^{2} a n d 17 x = x \times 0 + 8 x^{2} - 24 x + 72 [using (i)] = 4 (2 x^{2} - 6 x + 17) + 4 = 4 \times 0 + 4 = 4 [using (i)]$

$(i i) We have, x = 3 + 2 i \Rightarrow x - 3 = 2 i \Rightarrow (x - 3)^{2} = (2 i)^{2} \Rightarrow x^{2} + 3^{2} - 2 \times 3 \times x = - 4 \Rightarrow x^{2} + 9 - 6 x \times 4 = 0 \Rightarrow x^{2} - 6 x + 13 = 0 . . . (i) Now, x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44 = x^{2} (x^{2} - 6 x + 13) + 6 x^{2} - 13 x^{2} - 4 x^{3} + 4 x^{2} + 8 x + 44 (adding and subtracting 6 x^{3} a n d 13 x^{2}) = x^{2} \times 0 + 2 x^{3} - 9 x^{2} + 8 x + 44 [using (i)] = 2 x (x^{2} - 6 x + 13) + 12 x^{2} - 26 x - 9 x^{2} + 8 x + 44 (adding and subtracting 12 x^{2} a n d 26 x) = 2 x \times 0 + 3 x^{2} - 18 x + 44 = 3 (x^{2} - 6 x + 13) + 5 [using (i)] = 3 \times 0 + 5 = 5$

$(i i i) we have x = - 1 + i \sqrt{2} \Rightarrow x + 1 = i \sqrt{2} \Rightarrow (x + 1)^{2} = (i \sqrt{2})^{2} \Rightarrow x^{2} + 1 + 2 x = 2 \Rightarrow x^{2} + 2 x + 3 = 0 . . . . (i) Now, x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9 = x^{2} (x^{2} + 2 x + 3) + 2 x^{3} + 3 x^{2} + 4 x + 9 = x^{2} \times 0 + 2 x (x^{2} + 2 x + 3) - x^{2} - 2 x + 9 [using (i)] = 2 x \times 0 - (x^{2} + 2 x + 3) + 3 + 9 [using (i) and adding and subtracting 3] = - 0 + 3 + 9 [using (i)] = 12$

$(i v) we have, x = \frac{1 + i}{\sqrt{2}} \Rightarrow \sqrt{2 x} = 1 + i \Rightarrow {(\sqrt{2} x)}^{2} = (1 + i)^{2} (squaring both sides) \Rightarrow 2 x^{2} = 1^{2} + (i)^{2} + 2 \times 1 \times i = 1 - 1 + 2 i \Rightarrow 2 x^{2} = 2 i \Rightarrow x^{2} = i \Rightarrow (x^{2})^{2} = (i)^{2} (squaring both sides) \Rightarrow x^{4} = - 1 \Rightarrow x^{4} + 1 = 0 . . . . (i) Now x^{6} + x^{4} + x^{2} + 1 = x^{6} + x^{2} + x^{4} + 1 = x^{2} (x^{4} + 1) + 1 (x^{4} + 1) = x^{2} \times 0 + 1 \times 0 [using (i)] = 0$

$(v) x + 2 = - \sqrt{3} i \Rightarrow x^{2} + 4 x + 7 = 0 ∴ 2 x^{2} + 5 x^{3} + 7 x^{2} - x + 41 = (x^{2} + 4 x + 7) (2 x^{2} - 3 x + 5) + 6 = 0 \times (2 x^{2} - 3 x + 5) + 6 = 6$

Suggest Corrections

Similar questions

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

x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i

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

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

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

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

x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i

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

x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}} .

Identify polynomials in the following :

(i) f(x) = $4 x^{3} - x^{2}$ - 3x + 7

(ii) g(x) = $2 x^{3} - 3 x^{2} + \sqrt{x} - 1$

(iii) p(x) = $\frac{2}{7} x^{2} - \frac{7}{4} x + 9$

(iv) q(x) = $2 x^{2} - 3 x + \frac{4}{x} + 2$

(v) h(x) = $x^{4} - x^{\frac{3}{2}} + x - 1$

(vi) f(x) = 2 + $\frac{3}{x} + 4 x$

p (x) = x^{4} + 4 x^{3} + 7 x^{2} - 4

x + 1

x^{4} - 4 x^{3} + 3 x^{2} - 2 x + 1

x = 1 + i

Join BYJU'S Learning Program