Solve (m-1)^2+(sqrt{3}+1)^2=1

$$( m - 1 ) ^ { 2 } + ( \sqrt { 3 } + 1 ) ^ { 2 } = 1$$

Solve for m

$$m=\frac{-\left(\sqrt{2}+\sqrt{6}\right)\sqrt[4]{3}i+2}{2}\approx 1-2.542459757i$$

$$m=\frac{\left(\sqrt{2}+\sqrt{6}\right)\sqrt[4]{3}i+2}{2}\approx 1+2.542459757i$$

Steps by Finding Square Root

Subtracting \left(\sqrt{3}+1\right)^{2} from itself leaves 0.

$$\left(m-1\right)^{2}=1-\left(\sqrt{3}+1\right)^{2}$$

Subtract \left(\sqrt{3}+1\right)^{2} from 1.

$$\left(m-1\right)^{2}=-2\sqrt{3}-3$$

Take the square root of both sides of the equation.

$$m-1=i\sqrt{2\sqrt{3}+3} m-1=-i\sqrt{2\sqrt{3}+3}$$

Add 1 to both sides of the equation.

$$m-1-\left(-1\right)=i\sqrt{2\sqrt{3}+3}-\left(-1\right) m-1-\left(-1\right)=-\frac{\sqrt{2}+\sqrt{6}}{2}\sqrt[4]{3}i-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$m=i\sqrt{2\sqrt{3}+3}-\left(-1\right) m=-\frac{\sqrt{2}+\sqrt{6}}{2}\sqrt[4]{3}i-\left(-1\right)$$

Subtract -1 from i\sqrt{3+2\sqrt{3}}.

$$m=\frac{\sqrt{2}\sqrt[4]{3}i}{2}+\frac{\sqrt[4]{108}i}{2}+1$$

Subtract -1 from -i\sqrt[4]{3}\times \frac{\sqrt{6}+\sqrt{2}}{2}.

$$m=-\frac{\left(\sqrt{2}+\sqrt{6}\right)\sqrt[4]{3}i}{2}+1$$

The equation is now solved.

$$m=\frac{\sqrt{2}\sqrt[4]{3}i}{2}+\frac{\sqrt[4]{108}i}{2}+1 m=-\frac{\left(\sqrt{2}+\sqrt{6}\right)\sqrt[4]{3}i}{2}+1$$