Solve (1+2i)^2=(a+bi)(2-i)

$$( 1 + 2 i ) ^ { 2 } = ( a + b i ) ( 2 - i )$$

Solve for a

$$a=-2+i-ib$$

Steps for Solving Linear Equation

Calculate 1+2i to the power of 2 and get -3+4i.

$$-3+4i=\left(a+bi\right)\left(2-i\right)$$

Use the distributive property to multiply a+bi by 2-i.

$$-3+4i=\left(2-i\right)a+\left(2-i\right)bi$$

Multiply 2-i and i to get 1+2i.

$$-3+4i=\left(2-i\right)a+\left(1+2i\right)b$$

Swap sides so that all variable terms are on the left hand side.

$$\left(2-i\right)a+\left(1+2i\right)b=-3+4i$$

Subtract \left(1+2i\right)b from both sides.

$$\left(2-i\right)a=-3+4i-\left(1+2i\right)b$$

Multiply -1 and 1+2i to get -1-2i.

$$\left(2-i\right)a=-3+4i+\left(-1-2i\right)b$$

The equation is in standard form.

$$\left(2-i\right)a=\left(-1-2i\right)b+\left(-3+4i\right)$$

Divide both sides by 2-i.

$$\frac{\left(2-i\right)a}{2-i}=\frac{\left(-1-2i\right)b+\left(-3+4i\right)}{2-i}$$

Dividing by 2-i undoes the multiplication by 2-i.

$$a=\frac{\left(-1-2i\right)b+\left(-3+4i\right)}{2-i}$$

Divide -3+4i+\left(-1-2i\right)b by 2-i.

$$a=-2+i-ib$$

Solve for b

$$b=ia+\left(1+2i\right)$$

Steps for Solving Linear Equation

Calculate 1+2i to the power of 2 and get -3+4i.

$$-3+4i=\left(a+bi\right)\left(2-i\right)$$

Use the distributive property to multiply a+bi by 2-i.

$$-3+4i=\left(2-i\right)a+\left(2-i\right)bi$$

Multiply 2-i and i to get 1+2i.

$$-3+4i=\left(2-i\right)a+\left(1+2i\right)b$$

Swap sides so that all variable terms are on the left hand side.

$$\left(2-i\right)a+\left(1+2i\right)b=-3+4i$$

Subtract \left(2-i\right)a from both sides.

$$\left(1+2i\right)b=-3+4i-\left(2-i\right)a$$

Multiply -1 and 2-i to get -2+i.

$$\left(1+2i\right)b=-3+4i+\left(-2+i\right)a$$

The equation is in standard form.

$$\left(1+2i\right)b=\left(-2+i\right)a+\left(-3+4i\right)$$

Divide both sides by 1+2i.

$$\frac{\left(1+2i\right)b}{1+2i}=\frac{\left(-2+i\right)a+\left(-3+4i\right)}{1+2i}$$

Dividing by 1+2i undoes the multiplication by 1+2i.

$$b=\frac{\left(-2+i\right)a+\left(-3+4i\right)}{1+2i}$$

Divide -3+4i+\left(-2+i\right)a by 1+2i.

$$b=ia+\left(1+2i\right)$$