First slide

Algebra of complex numbers

Question

If $(a + b i)^{11} = x + i y,$ where $a, b, x, y \in R,$ then ${(b + a i)}^{11}$ equals

Moderate

A

$y + i x$
B

$- y - i x$
C

$- x - i y$
D

$x + i y$

Solution

$\bar{(a + b i)^{11}} = \bar{x + i y}$
$\begin{array}{l} \Rightarrow (a - b i)^{11} = x - i y \\ \Rightarrow [(- i) (b + i a)]^{11} = - i (y + i x) \\ \Rightarrow (- i)^{11} (b + i a)^{11} = - i (y + i x) \end{array}$
As $(- i)^{11} = (- i)^{8} (- 1)^{3} i^{3} = i,$
We get $(b + i a)^{11} = - (y + i x) = - y - i x$

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