Solve [(1-i)^2+(3+3i)^2]/1+i=a+bi

$$\frac { [ ( 1 - i ) ^ { 2 } + ( 3 + 3 i ) ^ { 2 } ] } { 1 + i } = a + b i$$

Solve for a

$$a=8+8i-ib$$

Solution Steps

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

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

Calculate 3+3i to the power of 2 and get 18i.

$$\frac{-2i+18i}{1+i}=a+bi$$

Add -2i and 18i to get 16i.

$$\frac{16i}{1+i}=a+bi$$

Multiply both numerator and denominator of \frac{16i}{1+i} by the complex conjugate of the denominator, 1-i.

$$\frac{16i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}=a+bi$$

Do the multiplications in \frac{16i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$\frac{16+16i}{2}=a+bi$$

Divide 16+16i by 2 to get 8+8i.

$$8+8i=a+bi$$

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

$$a+bi=8+8i$$

Subtract bi from both sides.

$$a=8+8i-bi$$

Multiply -1 and i to get -i.

$$a=8+8i-ib$$

Solve for b

$$b=ia+\left(8-8i\right)$$

Steps for Solving Linear Equation

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

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

Calculate 3+3i to the power of 2 and get 18i.

$$\frac{-2i+18i}{1+i}=a+bi$$

Add -2i and 18i to get 16i.

$$\frac{16i}{1+i}=a+bi$$

Multiply both numerator and denominator of \frac{16i}{1+i} by the complex conjugate of the denominator, 1-i.

$$\frac{16i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}=a+bi$$

Do the multiplications in \frac{16i\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$\frac{16+16i}{2}=a+bi$$

Divide 16+16i by 2 to get 8+8i.

$$8+8i=a+bi$$

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

$$a+bi=8+8i$$

Subtract a from both sides.

$$bi=8+8i-a$$

The equation is in standard form.

$$ib=8+8i-a$$

Divide both sides by i.

$$\frac{ib}{i}=\frac{8+8i-a}{i}$$

Dividing by i undoes the multiplication by i.

$$b=\frac{8+8i-a}{i}$$

Divide 8+8i-a by i.

$$b=ia+\left(8-8i\right)$$