If (1+i)x-i/2+i+(1+2i)y+i/2-i=1, then (x, y) isequal to

Correct Answer is : (73,−715)≤ft(7/3,-7/15)(37,15−7)

Correct Answer is : (73,−715)≤ft(7/3,-7/15)(37,15−7)

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AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

Question : 101 of 160

Marks: +1, -0

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

(x, y)

isequal to

Solution:

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

\Rightarrow\frac{[(1+i)x-i](2-i)}{(4-i^2)}+\frac{[(1+2i)y+i](2+i)}{(4-i^2)}

=1

⇒

\frac{2(1+i)x-2i-i(1+i)x+i^2}{4+1}

+\frac{2(1+2i)y+2i+i(1+2i)y+i^2}{(4+1)}=1

\Rightarrow\frac{(2+2i-i-i^2)x-2i+i^2}{5}

+\frac{(4i+2+i+2i^2)y+2i+i^2}{5}=1

\Rightarrow\frac{(2+i+1)x-2i-1}{5}+\frac{(5i+2-2)y+2i-1}{5}=1

\Rightarrow(3+i)x-2i-1+(5i)y

+2i-1=5

\Rightarrow(3+i)x+5iy=7

\Rightarrow3x+ix+5iy-7=0

\Rightarrow(3x-7)+(x+5y)i=0+i0

On comparing, we get

3x-7=0

⇒

x=\frac{7}{3}

and

x+5y=0

⇒

y=\frac{-7}{15}

Hence,

(x,y)=\left(\frac{7}{3},\frac{-7}{15}\right)

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