Question

Find the value of $x^2+y^2$ for which the complex numbers $-3-i{x}^{2}y$ and $x^2+y+4i$ are equal , where x and y are real numbers .

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Answer 1

Given two complex number z₁ and z₂ which are equal

$-3-i{x}^{2}y$ = $x^2+y+4i$

Comparing real and imaginary part on both sides

$x^2+y$ = -3 - - - - - - - - (I)

$-x^{2}y$ = 4 - - - - - - - - (II)

$x^2$ = $\frac{-4}{y}$

Substitute the value of $x^2$ in equation (1).

$\frac{-4}{y}$ + y = -3

$y^2$ + 3y - 4 = 0

(y + 4)(y - 1) = 0

y = -4, y = 1

Substitute the value of y in equation (2)

When y = -4

$-x^2$ (-4) = 4

$x^2$ = 1, x = $\pm 1$

when y = 1

$-x^2$(I) = 4

$x^2$ = -4 (not possible)

We won't get real number

So, x = $\pm 1$, y = -4

So, $x^2+y^2$ = 1 + 16 = 17