Solve (sqrt{3}+i)^10=a+ib

$$( \sqrt { 3 } + i ) ^ { 10 } = a + i b$$

Solve for a

$$a=-ib-512\sqrt{3}i+512$$

Solution Steps

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

$$a+ib=\left(\sqrt{3}+i\right)^{10}$$

Subtract ib from both sides.

$$a=\left(\sqrt{3}+i\right)^{10}-ib$$

Reorder the terms.

$$a=-ib+\left(\sqrt{3}+i\right)^{10}$$

Solve for b

$$b=ia-512\sqrt{3}-512i$$

Steps for Solving Linear Equation

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

$$a+ib=\left(\sqrt{3}+i\right)^{10}$$

Subtract a from both sides.

$$ib=\left(\sqrt{3}+i\right)^{10}-a$$

The equation is in standard form.

$$ib=-a+\left(\sqrt{3}+i\right)^{10}$$

Divide both sides by i.

$$\frac{ib}{i}=\frac{-a-512\sqrt{3}i+512}{i}$$

Dividing by i undoes the multiplication by i.

$$b=\frac{-a-512\sqrt{3}i+512}{i}$$

Divide 512-512i\sqrt{3}-a by i.

$$b=ia-512\sqrt{3}-512i$$