Question #350645

Find the four roots of z⁴=-1


1
Expert's answer
2022-06-15T05:49:22-0400

z4=1z^4=-1

z=14z=\sqrt[4]{-1}


The roots of a complex number are also given by a formula zn=zn(cosφ+2πkn+isinφ+2πkn)\sqrt[n]{z}=\sqrt[n]{|z|}\left(\cos\frac{\varphi+2\pi k}{n}+i\sin\frac{\varphi+2\pi k}{n}\right), where k=0,n1k=\overline{0,n-1}.


1=1|-1|=1, φ=π\varphi=\pi


k=0k=0:

z1=14=14(cosπ+2π04+isinπ+2π04)=1(cosπ4+isinπ4)=22+i22z_1=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot0}{4}+i\sin\frac{\pi+2\pi\cdot0}{4}\right)=1\cdot\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}


k=1k=1:

z2=14=14(cosπ+2π14+isinπ+2π14)=1(cos3π4+isin3π4)=22i22z_2=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot1}{4}+i\sin\frac{\pi+2\pi\cdot1}{4}\right)=1\cdot\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}


k=2k=2:

z3=14=14(cosπ+2π24+isinπ+2π24)=1(cos5π4+isin5π4)=22i22z_3=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot2}{4}+i\sin\frac{\pi+2\pi\cdot2}{4}\right)=1\cdot\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}


k=3k=3:

z4=14=14(cosπ+2π34+isinπ+2π34)=1(cos7π4+isin7π4)=22+i22z_4=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot3}{4}+i\sin\frac{\pi+2\pi\cdot3}{4}\right)=1\cdot\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}


So z1=22+i22z_1=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}, z2=22i22z_2=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}, z3=22i22z_3=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}, z4=22+i22z_4=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}


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