Real values of x and y for which equation ({x^4} + 2xi) - (3{x^2} + yi) = (3

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4-1.Complex numbers

medium

The real values of $x$ and $y$ for which the equation $({x^4} + 2xi) - (3{x^2} + yi) = $$(3 - 5i) + (1 + 2yi)$ is satisfied, are

$x = 2,y = 3$

$x = - 2,y = \frac{1}{3}$

Both $(a)$ and $(b)$

None of these

Solution

(c) Given equation
$({x^4} + 2xi) – (3{x^2} + yi) = (3 – 5i) + (1 + 2yi)$
$ \Rightarrow \,\,\,({x^4} – 3{x^2}) + i(2x – 3y) = 4 – 5i$
Equating real and imaginary parts, we get
${x^4} – 3{x^2} = 4$ ……$(i)$
and $2x – 3y = – 5$ …..$(ii)$
From $(i)$ and $(ii),$ we get $x = \pm 2$and $y = 3,\frac{1}{3}$
Trick : Put $x = 2,y = 3$and then $x = – 2,$$y = \frac{1}{3},$ we see that they both satisfy the given equation.

Standard 11

Mathematics

The value of ${(1 + i)^8} + {(1 – i)^8}$ is

easy

If $z_1, z_2, z_3$ are the roots of the equation $z^3 – z^2(4 + 3i) + z(3 + 8i) – 5i = 0$, then $Re(z_1) + Re(z_2) + Re(z_3)$ is –

normal

${i^2} + {i^4} + {i^6} + ……$upto $(2n + 1)$ terms =