Asked by putman/raj
plz help find the fourth root of -16,giving the result in form of a+jb
and also 5th root of -1,giving the result in polar form plz help thanks
and also 5th root of -1,giving the result in polar form plz help thanks
Answers
Answered by
Steve
?-16 = ?(16 cis ?) = ?16 cis ?/4 = 2(1/?2 + 1/?2 i) = ?2 + ?2 i
That is the 1st root. There is an equivalent root in each quadrant. See
http://www.wolframalpha.com/input/?i=x%5E4+%3D+-16
Work the 5th root the same way, using de Moivre's theorem.
That is the 1st root. There is an equivalent root in each quadrant. See
http://www.wolframalpha.com/input/?i=x%5E4+%3D+-16
Work the 5th root the same way, using de Moivre's theorem.
Answered by
putman/raj
sir how did u get the argument? And the modulus?
Putoyanah
Putoyanah
Answered by
bobpursley
here is another way, to answer your argument and modulus question.
-1=1@(180+n*360
fifth root of that is 1@36+n(72)
so the roots are
1@36
1@108
1@180
and you can do the last two. Now that angle is on the complex plane, If you want it in real,imag format, it is converted by
real=1*cosTheta+i*1*sinTheta
if you convert all those, you will see a pattern of symmetry, clockwise and counterclockwise.
-1=1@(180+n*360
fifth root of that is 1@36+n(72)
so the roots are
1@36
1@108
1@180
and you can do the last two. Now that angle is on the complex plane, If you want it in real,imag format, it is converted by
real=1*cosTheta+i*1*sinTheta
if you convert all those, you will see a pattern of symmetry, clockwise and counterclockwise.
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