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Question: The values of \(x,y \in R\) and \((x + iy)(3 + 2i) = 1 + i\) for which the numbers \((x,y)\) and \(\...

The values of x,yRx,y \in R and (x+iy)(3+2i)=1+i(x + iy)(3 + 2i) = 1 + i for which the numbers (x,y)(x,y) and (1,15)\left( 1,\frac{1}{5} \right) are conjugate complex can be.

A

(113,113)\left( \frac{1}{13},\frac{1}{13} \right)or (513,113)\left( \frac{5}{13},\frac{1}{13} \right)

B

(15,15)\left( \frac{1}{5},\frac{1}{5} \right)or (1i1+i)100=a+ib\left( \frac{1 - i}{1 + i} \right)^{100} = a + ib

C

a=2,b=1a = 2,b = - 1or a=1,b=0a = 1,b = 0

D

None of these

Answer

(113,113)\left( \frac{1}{13},\frac{1}{13} \right)or (513,113)\left( \frac{5}{13},\frac{1}{13} \right)

Explanation

Solution

According to condition,

726i25\frac{- 7 - 26i}{25}

\becauseand a=aa = \overline{a}(z+a)(z+a)=(z+a)(z+a)=(z+a)(z+a)(z + a)(\overline{z} + a) = (z + a)(\overline{z} + \overline{a}) = (z + a)(\overline{z + a})

=z+a2= |z + a|^{2}or ziz+i=x+i(y1)x+i(y+1).xi(y+1)xi(y+1)\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)}.\frac{x - i(y + 1)}{x - i(y + 1)}