(-1+0i)^{\frac{1}{4}} &amp;= [1(\cos (\pi+2\pi k)+i \sin (\pi+2\pi k))]^{\frac{1}{4}}\\(-1+0i)^{\frac{1}{4}} &amp;= 1^{\frac{1}{4}} \left(\cos \frac{\pi+2\pi k}{4}+i \sin \frac{\pi+2\pi k}{4}\right)\\x_1 &amp;= 1 \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} &amp;&amp; \text{for} \ k=0\\x_2 &amp;= 1 \left(\cos \frac{3\pi}{4}+i \sin \frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} &amp;&amp; \text{for} \ k=1\\x_3 &amp;= 1 \left(\cos \frac{5\pi}{4}+i \sin \frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} &amp;&amp; \text{for} \ k=2\\x_4 &amp;= 1 \left(\cos \frac{7\pi}{4}+i \sin \frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} &amp;&amp; \text{for} \ k=3