First slide

Algebra of complex numbers

Question

The real values of x and y for which the following equality hold, are respectively $(x^{4} + 2 xi) - (3 x^{2} + iy) = (3 - 5 i) + (1 + 2 iy)$

Moderate

A

$2, 3 or - 2, 1 / 3$
B

$1, 3 or - 1, 1 / 3$
C

$2, 1 / 3 or - 2, 3$
D

None of these

Solution

The given equality can be rewritten as
$\begin{array}{ll} (x^{4} - 3 x^{2}) + i (2 x - y) = 4 + i (2 y - 5) \\ \Rightarrow & x^{4} - 3 x^{2} = 4 and 2 x - y = 2 y - 5 \\ \Rightarrow & x^{4} - 3 x^{2} - 4 = 0 and 2 x - 3 y + 5 = 0 \\ \Rightarrow & (x^{2} - 4) (x^{2} + 1) = 0 [∵ x^{2} \neq - 1] \\ \Rightarrow & x = \pm 2 \end{array}$
$∴ At x = 2, y = 3 and at x = - 2, y = \frac{1}{3}$

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