Answer to Question #107109 in Algebra for Chinmoy Kumar Bera

Question

obtain the 6th root of -7 and represent them in argand diagram

Answer 1

$\sqrt[6]{-7}$ we need to write the number -7 in trigonometric form: $-7=7(cos π + і sin π)$ $z_k=\sqrt[6]{7}(cos⁡(π/6+(2π/6) k)+i sin⁡(π/6+(2π/6) k)), k=0,1,2,3,4,5$

from here

$z_0=\sqrt[6]{7}(cos⁡(π/6)+i sin⁡(π/6))=(\sqrt[6]{7}\sdot\sqrt{3})/2+(\sqrt[6]{7}/2)i\\$ ;

$z_1=\sqrt[6]{7}(cos⁡(π/2)+i sin⁡(π/2))=\sqrt[6]{7}i\\$ ;

$z_2=\sqrt[6]{7}(cos⁡(5π/6)+i sin⁡(5π/6))=-(\sqrt[6]{7}\sdot\sqrt{3})/2+(\sqrt[6]{7}/2)i\\$ ;

$z_3=\sqrt[6]{7}(cos⁡(5π/6)+i sin⁡(5π/6))=-(\sqrt[6]{7}\sdot\sqrt{3})/2-(\sqrt[6]{7}/2)i\\$;

$z_4=\sqrt[6]{7}(cos⁡(3π/2)+i sin⁡(3π/2))=-\sqrt[6]{7}i\\$;

$z_5=\sqrt[6]{7}(cos⁡(5π/6)+i sin⁡(5π/6))=(\sqrt[6]{7}\sdot\sqrt{3})/2-(\sqrt[6]{7}/2)i\\$ ;

Answer: $z_0=(\sqrt[6]{7}\sdot\sqrt{3})/2+(\sqrt[6]{7}/2)i;$

$z_1=\sqrt[6]{7}i$ ;

$z_2=-(\sqrt[6]{7}\sdot\sqrt{3})/2+(\sqrt[6]{7}/2)i$;

$z_3=-(\sqrt[6]{7}\sdot\sqrt{3})/2-(\sqrt[6]{7}/2)i$;

$z_4=-\sqrt[6]{7}i$;

$z_5=(\sqrt[6]{7}\sdot\sqrt{3})/2-(\sqrt[6]{7}/2)i$.