If 1- i 2-2-1- x - iy- find x - y

(1+i)^2/2 i=x+iy ⇒ (1+2i 1)/2 i=x+iy ⇒ 2i/2 i=x+iy ⇒ 2i(2+i)/(2 i)(2+i)=x+iy [Rationalizing the denominator] ⇒ 2(2i 1)/4+1=x+iy ⇒ 4i 2/5=x+iy ⇒ 2/5+i4 ...

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Byju's Answer

Standard XII

Mathematics

Equality of 2 Complex Numbers

If 1+ i 2/2-1...

Question

$If \frac{(1 + i)^{2}}{2 - i} = x + i y, find x + y .$

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Solution

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

$\Rightarrow \frac{(1 + 2 i - 1)}{2 - i} = x + i y \Rightarrow \frac{2 i}{2 - i} = x + i y \Rightarrow \frac{2 i (2 + i)}{(2 - i) (2 + i)} = x + i y [Rationalizing the denominator] \Rightarrow \frac{2 (2 i - 1)}{4 + 1} = x + i y \Rightarrow \frac{4 i - 2}{5} = x + i y \Rightarrow - \frac{2}{5} + i \frac{4}{5} = x + i y$

Comparing the real and imaginary parts, we get

$x = - \frac{2}{5}, y = \frac{4}{5} x + y = \frac{2}{5}$

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If (1+i)22−i=x+iy, find x+y.

(1+i)22−i=x+iy ⇒ (1+2i−1)2−i=x+iy⇒ 2i2−i=x+iy⇒ 2i(2+i)(2−i)(2+i)=x+iy [Rationalizing the denominator]⇒ 2(2i−1)4+1=x+iy⇒ 4i−25=x+iy ⇒ −25+i45=x+iy Comparing the real and imaginary parts, we get x=−25,y=45x+y=25

$If \frac{(1 + i)^{2}}{2 - i} = x + i y, find x + y .$