Solve sqrt{1+k^{2}}sqrt{(-4km/2k^2+1)^2-4(2m^2-2/2k^2+1)}=2

$$\sqrt { 1 + k ^ { 2 } } \sqrt { ( \frac { - 4 k m } { 2 k ^ { 2 } + 1 } ) ^ { 2 } - 4 ( \frac { 2 m ^ { 2 } - 2 } { 2 k ^ { 2 } + 1 } ) } = 2$$

Solve for k (complex solution)

$$k=-\frac{i\left(1-m^{2}\right)^{-\frac{1}{2}}\sqrt{2-4m^{2}}}{2}$$

$$k=\frac{i\left(1-m^{2}\right)^{-\frac{1}{2}}\sqrt{2-4m^{2}}}{2}\text{, }m\neq 1\text{ and }m\neq -1\text{ and }m\neq 0$$

Solve for m (complex solution)

$$m=-\frac{i\left(k^{2}+1\right)^{-\frac{1}{2}}\sqrt{-4k^{2}-2}}{2}$$

$$m=\frac{i\left(k^{2}+1\right)^{-\frac{1}{2}}\sqrt{-4k^{2}-2}}{2}\text{, }k\neq -\frac{\sqrt{2}i}{2}\text{ and }k\neq \frac{\sqrt{2}i}{2}\text{ and }k\neq -i\text{ and }k\neq i$$

Solve for k

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

$$k=-\frac{\sqrt{\frac{2\left(2m^{2}-1\right)}{1-m^{2}}}}{2}\text{, }\left(m\geq \frac{\sqrt{2}}{2}\text{ and }m<1\right)\text{ or }\left(m>-1\text{ and }m\leq -\frac{\sqrt{2}}{2}\right)$$

Solve for m

$$m=\frac{\sqrt{\frac{2\left(2k^{2}+1\right)}{k^{2}+1}}}{2}$$

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