If x-i y-a-i b-c-i d- then x2-y22- then is equal toA- a-bc-d-a-bc-da2-b2B- a2-b2-c2-d2D- a2-b2-c2-d2

The correct option is B x+iy = √(a+ib/c+id) ⇒ |x+iy| = √( |a+ib/c+id |) ⇒ x^2 + y^2 = √(a^2 + b^2/c^2 + d^2) ⇒ (x^2 + y^2)^2 = a^2 + b^2/c^2 + d^2 ...

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Byju's Answer

Standard XI

Mathematics

Progression

If x+i y=√a+i...

Question

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

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Solution

The correct option is A
$x + i y = \sqrt{\frac{a + i b}{c + i d}} \Rightarrow | x + i y | = \sqrt{∣ ∣ \frac{a + i b}{c + i d} ∣ ∣} \Rightarrow x^{2} + y^{2} = \sqrt{\frac{a^{2} + b^{2}}{c^{2} + d^{2}}} \Rightarrow (x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}$

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Similar questions

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}}

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}}

Q. If

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

, then

\frac{a^{2} + b^{2}}{c^{2} + d^{2}}

is equal to

If $x - i y = \sqrt{\frac{a - i d}{c - i d}},$ prove that $(x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}$ .

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If x+iy=√a+ibc+id,then (x2+y2)2, then is equal to

The correct option is A x+iy=√a+ibc+id⇒|x+iy|=√∣∣a+ibc+id∣∣⇒x2+y2=√a2+b2c2+d2⇒(x2+y2)2=a2+b2c2+d2

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

The correct option is A
$x + i y = \sqrt{\frac{a + i b}{c + i d}} \Rightarrow | x + i y | = \sqrt{∣ ∣ \frac{a + i b}{c + i d} ∣ ∣} \Rightarrow x^{2} + y^{2} = \sqrt{\frac{a^{2} + b^{2}}{c^{2} + d^{2}}} \Rightarrow (x^{2} + y^{2})^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}$