F a - ib - c - id - x - iy- then a 2- b 2- c 2- d 2 is equal toA- x-y2B- x-y2C- x2-y2D- x2-y2

Given : a+ibc+id=x+iy Now, nbsp; a^2+b^2c^2+d^2 = (a+ib)(a ib)(c+id)(c id) nbsp;= a+ibc+id×a ibc id nbsp;= a+ibc+id×a+ibc+id nbsp;= (x+iy)× (x iy) = ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

f a + ib / c ...

Question

If $\frac{a + i b}{c + i d} = x + i y$ , then $\frac{a^{2} + b^{2}}{c^{2} + d^{2}}$ is equal to

x^{2} - y^{2}

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x^{2} + y^{2}

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(x + y)^{2}

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(x - y)^{2}

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Solution

The correct option is B $x^{2} + y^{2}$
Given : $\frac{a + i b}{c + i d} = x + i y$
Now,
$\frac{a^{2} + b^{2}}{c^{2} + d^{2}} = \frac{(a + i b) (a - i b)}{(c + i d) (c - i d)} = \frac{a + i b}{c + i d} \times \frac{a - i b}{c - i d} = \frac{a + i b}{c + i d} \times \frac{¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ a + i b}{¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ c + i d} = (x + i y) \times (x - i y) = x^{2} + y^{2}$

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Similar questions

I f x + i y = \sqrt{\frac{a + i b}{c + i d}}, t h e n (x^{2} + y^{2})^{2}

, then is equal to

Q. If

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

Show that,
i)

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

ii)

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

Q. If

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

prove that

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

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

, prove that

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

Q. Let

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

, then prove that

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

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If a+ibc+id=x+iy, then a2+b2c2+d2 is equal to

The correct option is B x2+y2Given : a+ibc+id=x+iy Now, a2+b2c2+d2=(a+ib)(a−ib)(c+id)(c−id)=a+ibc+id×a−ibc−id=a+ibc+id×¯¯¯¯¯¯¯¯¯¯¯¯¯¯a+ib¯¯¯¯¯¯¯¯¯¯¯¯¯¯c+id=(x+iy)×(x−iy)=x2+y2

If $\frac{a + i b}{c + i d} = x + i y$ , then $\frac{a^{2} + b^{2}}{c^{2} + d^{2}}$ is equal to