# Find x + iy: 4 + 2i + ( 1 – i)/ ( x + iy) – (9 + 7i)(3 + 2i) = (32 + i)

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### 2 Answers

We have to find x + iy given: 4 + 2i + (1 – i)/ (x + iy) – (9 + 7i) (3 + 2i) = (32 + i)

4 + 2i + (1 – i)/ (x + iy) – (9 + 7i) (3 + 2i) = (32 + i)

=> 4 + 2i + (1 – i)/ (x + iy) – 27 – 21i – 18i – 14i^2 = (32 + i)

=> 4 – 27 + 14 + 2i – 21i – 18i + (1 – i)/ (x + iy) = 32 + i

=> -9 – 37i + (1 – i)/ (x + iy) = 32 + i

=> (1 – i)/ (x + iy) = 32 + 9 + i + 37i

=> (1 – i)/ (x + iy) = 41 + 38i

=> x + iy = (1 – i)/ (41 + 38i)

=> x + iy = (1 – i) (41 – 38i)/ (41^2 + 38^2)

=> x + iy = (41 – 38i – 41i + 38i^2) / 3125

=> x + iy = (3 – 79i)/3125

**Therefore x + iy = 3/3125 – 79i/3125**

To find x + iy: 4 + 2i + ( 1 – i)/ ( x + iy) – (9 + 7i)(3 + 2i) = 32 + i.

We rewrite the given expression as:

( 1 – i)/ ( x + iy) = (9 + 7i)(3 + 2i) + 32+i-4-2i

(1-i)/(x+iy) = (9+7i)(3+2i)+ 28 - i.

(1-i)/(x-yi) = (27+18i+21i+14i^2)+28-i

(1-i)/(x+yi) = (13 +39i)+28-i, as i^2= -1.

(1-i)/(x+yi) = 41+38i

(1-i)(x-yi)/(x+y^2-y^2*i^2). = 41+38i

(x-y)/(x^2+y^2) - (x+y)i/(x^2+y^2) = 41+38i(1)

We equate real parts on both sides of (1) and the equate imaginary parts on both sides separately:

Real parts: (x-y)/(x^2+y^2) = 41....(2)

Imaginary parts: (x+y)/(x^2+y^2) = -38....(3)

(2)/(3): (x-y)/(x+y) = 41/-38

38(x-y) = -41(x+y).

(38+41)x = (-41+38)y

79x = -3y.

x = -(3/79)y

Substitute x= -3y/79 in (x+y)/(x^2+y^2) = -38:

(-3/79+1)y/{(-3/79)^2 +1}y^2 = -38.

79(76)y/{3^2+79^2)y^2 = -38.

76*79y = -38(3^2+79^2)y^2

y = 76*79/(-38)(6250) = -79/3125.

Therefore x = -3y/79 = -3*-79/79*3125 = 3/3125.

Therefore x = 3/3125 and y = -79/3125.

So x+yi = (3/3125)+(-79/3125)i.