Question

if x+iy=√1+i/1-i prove that x^2+y^2=1​

Answer 1

Step-by-step explanation:

⇒ x + iy = $\sqrt{\frac{1+i}{1-i}}$

             = $\sqrt{\frac{1+i}{1-i}\times\frac{1+i}{1+i}}$

             = $\sqrt{\frac{(1+i)^2}{(1-i)(1+i)}}$

             = $\sqrt{\frac{1+i^2+2(1)(i)}{1-i^2}}$

             = $\sqrt{\frac{1-1+2i}{1-(-1)}}$

             = $\sqrt{\frac{2i}{1+1}}$

             = $\sqrt{\frac{2i}{2}}$

             =$\sqrt{1i}$

Taking modulus both sides:

⇒ | x + iy | = | √1i  |

⇒ x² + y ² = √1²

⇒ x² + y² = 1  

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ x + iy = \sqrt{ \frac{1 + i}{1 - i} } }$

TO PROVE

$\displaystyle \sf{ {x}^{2} + {y}^{2} = 1 }$

EVALUATION

Here the given equation equation is

$\displaystyle \sf{ x + iy = \sqrt{ \frac{1 + i}{1 - i} } }$

$\displaystyle \sf{ \implies \: x + iy = \sqrt{ \frac{ - {i}^{2} + i}{1 - i} } }$

$\displaystyle \sf{ \implies \: x + iy = \sqrt{ \frac{ i(1 - i)}{1 - i} } }$

$\displaystyle \sf{ \implies \: x + iy = \sqrt{i} }$

Squaring both sides we get

$\displaystyle \sf{ \implies \: {x}^{2} + 2ixy + {i}^{2} {y}^{2} = i}$

$\displaystyle \sf{ \implies \: {x}^{2} + 2ixy - {y}^{2} = i}$

$\displaystyle \sf{ \implies \: {x}^{2} - {y}^{2} + 2ixy = i}$

Comparing both sides we get

$\displaystyle \sf{ \implies \: {x}^{2} - {y}^{2} = 0 \: \: \: and \: \: \: 2xy = 1}$

Now

$\displaystyle \sf{ {( {x}^{2} + {y}^{2} )}^{2} }$

$\displaystyle \sf{ = {( {x}^{2} - {y}^{2} )}^{2} +4 {x}^{2} {y}^{2} }$

$\displaystyle \sf{ = {(0)}^{2} + {(2xy)}^{2} }$

$\displaystyle \sf{ = {(1)}^{2} }$

$\displaystyle \sf{ = 1}$

$\displaystyle \sf{ \implies {( {x}^{2} + {y}^{2} )}^{2} = 1}$

$\displaystyle \sf{ \implies {( {x}^{2} + {y}^{2} )}^{} = 1}$

Hence proved

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