Answer to Question #350645 in Complex Analysis for John

Question #350645

Find the four roots of z⁴=-1


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

"z^4=-1"

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


The roots of a complex number are also given by a formula "\\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=\\overline{0,n-1}".


"|-1|=1", "\\varphi=\\pi"


"k=0":

"z_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=1":

"z_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=2":

"z_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=3":

"z_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 "z_1=\\frac{\\sqrt{2}}{2}+i\\frac{\\sqrt{2}}{2}", "z_2=\\frac{\\sqrt{2}}{2}-i\\frac{\\sqrt{2}}{2}", "z_3=-\\frac{\\sqrt{2}}{2}-i\\frac{\\sqrt{2}}{2}", "z_4=-\\frac{\\sqrt{2}}{2}+i\\frac{\\sqrt{2}}{2}"


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