Find the real part of x - iy - i - x - iy -2Take - x -2- iy -1- k A- k-x-2 x-y-1 y- k-x-1 x-y-2 y- k-xx-2-yy-1 y-D- k-xx-1-yy-2 y-

(x + iy) + i/(x + iy) + 2 = x + (y+1)i/(x + 2) + iy = x + i (y + 1)/(x + 2) + iy×(x+2) iy/(x + 2) iy ⇒ Real part = (x+2)x + (y + 1)y/(x + 2)^2 y^2 = k[(x ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

Find the real...

Question

Find the real part of $(\frac{(x + i y) + i}{(x + i y) + 2})$
Take $(| (x + 2) + i y |)$ = $\sqrt{\frac{1}{k}}$

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Solution

The correct option is A

$\frac{(x + i y) + i}{(x + i y) + 2}$ = $\frac{x + (y + 1) i}{(x + 2) + i y}$
= $\frac{x + i (y + 1)}{(x + 2) + i y} \times \frac{(x + 2) - i y}{(x + 2) - i y}$
$\Rightarrow$ Real part = $\frac{(x + 2) x + (y + 1) y}{(x + 2)^{2} - y^{2}}$
= $k [(x + 2) x + (y + 1) y]$

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Find the real part of ((x+iy)+i(x+iy)+2) Take (|(x+2)+iy|) =√1k

The correct option is A (x+iy)+i(x+iy)+2 = x+(y+1)i(x+2)+iy = x+i(y+1)(x+2)+iy×(x+2)−iy(x+2)−iy ⇒ Real part = (x+2)x+(y+1)y(x+2)2−y2 = k[(x+2)x+(y+1)y]

Find the real part of $(\frac{(x + i y) + i}{(x + i y) + 2})$
Take $(| (x + 2) + i y |)$ = $\sqrt{\frac{1}{k}}$

The correct option is A

$\frac{(x + i y) + i}{(x + i y) + 2}$ = $\frac{x + (y + 1) i}{(x + 2) + i y}$
= $\frac{x + i (y + 1)}{(x + 2) + i y} \times \frac{(x + 2) - i y}{(x + 2) - i y}$
$\Rightarrow$ Real part = $\frac{(x + 2) x + (y + 1) y}{(x + 2)^{2} - y^{2}}$
= $k [(x + 2) x + (y + 1) y]$