What is the answer to (2+3i)*(x+iy)=(2+i)^2 ?

The answer to (2+3i)*(x+iy)=(2+i)^2 is x= 18/13 ,y=-1/13

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(2+3i)*(x+iy)=(2+i)^2

(2 + 3 i) \cdot (x + i y) = (2 + i)^{2}

x = \frac{18}{13}, y = - \frac{1}{13}

Solution steps

(2 + 3 i) (x + i y) = (2 + i)^{2}

Expand

(2 x - 3 y) + i (2 y + 3 x) = 3 + 4 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[2 x - 3 y = 3 2 y + 3 x = 4]

[2 x - 3 y = 3 2 y + 3 x = 4] : x = \frac{18}{13}, y = - \frac{1}{13}

x = \frac{18}{13}, y = - \frac{1}{13}

∣ z - (1 + i) ∣ = 1

a + bi = \frac{2 i}{1 + i}

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

(x + i y) (3 - 2 i) = 4 + i

x, x sin (y) - x^{2} y^{- 3} = c

What is the answer to (2+3i)*(x+iy)=(2+i)^2 ?
The answer to (2+3i)*(x+iy)=(2+i)^2 is x= 18/13 ,y=-1/13