If x- iy -a- id -c-i d- prove that x2-y22-a2-b2-c2-d2-

Here x iy=√(a id/c id) Squaring both sides, we get (x iy)^2=a ib/c id ⇒ |(x iy)^2|= | a ib/c id | ⇒ |x iy||x iy|=|a ib|/|c id| ⇒ (√(x^2+y^2))(√(x^2+y^2))=√(a^2+ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Conjugate of a Complex Number

If x- iy =√a-...

Question

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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Solution

Here $x - i y = \sqrt{\frac{a - i d}{c - i d}}$

Squaring both sides, we get

$(x - i y)^{2} = \frac{a - i b}{c - i d}$

$\Rightarrow | (x - i y)^{2} | = ∣ ∣ \frac{a - i b}{c - i d} ∣ ∣$

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

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

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

Squaring both sides

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

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Similar questions

Q. If

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. If

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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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}

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If x−iy=√a−idc−id, prove that (x2+y2)2=a2+b2c2+d2.

Here x−iy=√a−idc−id Squaring both sides, we get (x−iy)2=a−ibc−id ⇒ |(x−iy)2|=∣∣a−ibc−id∣∣ ⇒ |x−iy||x−iy|=|a−ib||c−id| ⇒ (√x2+y2)(√x2+y2)=√a2+b2√c2+d2 ⇒ (x2+y2)=√a2+b2c2+d2 Squaring both sides (x2+y2)2=a2+b2c2+d2

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