How do I solve z^{5}=-2i (complex numbers)?

sunshine022uv

sunshine022uv

Answered question

2022-01-16

How do I solve z5=2i (complex numbers)?

Answer & Explanation

sirpsta3u

sirpsta3u

Beginner2022-01-17Added 42 answers

z5=2i Convert the right side into exponential form z5=2eiπ2+2kiπ z=25eπ10+2kiπ5 z=25[cos(4k+1)π10+isin(4k+1)π10] Putting k=0,1,2,3,4... z1=25[cosπ10+isinπ10] z2=25[cosπ2+isinπ2]=25i z3=25[cos9π10+isin9π10] z4=25[cosπ10isinπ10] z5=25[cos9π10isin9π10]
levurdondishav4

levurdondishav4

Beginner2022-01-18Added 38 answers

I would use polar coordinates, as extracting roots becomes far easier.
2i=2e3π2
z=[215]e3π10+2jπ5 where j=0,1,2,3,or 4 to get the 5 roots of this equation.
I used the fact that [eix]15=eix5. This comes from the identity for exponents that (ea)b=eab.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

2i=2(eip/2) therefore z=20.2(eip/2)15 where,p=πe and i=1; now by applying de morvies

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