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Algebra of Complex Numbers

a, b, c, d ...

Question

a, b, c, d $\in$ R are such that $a^{2} + b^{2} = 4$ and $c^{2} + d^{2} = 2$ and if $(a + i b)^{2} = (c + i d)^{2} (x + i y)$ then $x^{2} + y^{2} =$ ?

A

4

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B

3

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C

2

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D

1

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Solution

The correct option is A $4$
Given $a^{2} + b^{1} = 4, c^{2} + d^{2} = 2$
and also
$(a + i b)^{2} = (c + i d)^{2} (x + i y)$
Applying modulus on both sides
$(| a + i b |)^{2} = (| c + i d |)^{2} (| x + i y |)$
$⟹ (a^{2} + b^{2}) = (c^{2} + d^{2}) \sqrt{x^{2} + y^{2}}$
$⟹ \sqrt{x^{2} + y^{2}} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}} = \frac{4}{2} = 2$
$⟹ \sqrt{x^{2} + y^{2}} = 2$
$⟹ x^{2} + y^{2} = 4$

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Similar questions

a, b, c, d ϵ R

a^{2} + b^{2} = 4

c^{2} + d^{2} = 2

| a + i b |^{2} = | c + i d |^{2} | x + i y |

x^{2} + y^{2} =

x - i y = \frac{a - i b}{c - i d}

x^{2} + y^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}

\in R^{+}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

\frac{(a + i b)}{(c + i d)} = x + i y

\frac{a - i b}{c - i d} = x - i y

\frac{a^{2} + b^{2}}{c^{2} + d^{2}} = x^{2} + y^{2}

x + i y = \sqrt{\frac{a + i b}{c + i d}}

(x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}

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