Find the real values of x and y, if (1 + i) (x + iy) = 2 – 5i

Asked by Sakshi | 1 year ago |  53

##### Solution :-

(1 + i) (x + iy) = 2 – 5i

Given:

(1 + i) (x + iy) = 2 – 5i

Divide with (1+i) on both the sides we get,

(x + iy) = $$\dfrac{ (2 – 5i)}{(1+i)}$$

Multiply and divide by (1-i)

= $$\dfrac{2 – 7i + 5(-1) }{ 2}$$ [since, i= -1]

= $$\dfrac{ (-3-7i)}{2}$$

Now, equating Real and Imaginary parts on both sides we get

x = $$\dfrac{-3}{2}$$ and y = $$\dfrac{-7}{2}$$

Thee real values of x and y are $$\dfrac{-3}{2}, \dfrac{-7}{2}$$

Answered by Aaryan | 1 year ago

### Related Questions

#### Show that 1 + i10 + i20 + i30 is a real number?

Show that 1 + i10 + i20 + i30 is a real number?

#### Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0

Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0

#### Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0

Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0