hello, how do i solve : (Z-2i)^4 = -8-8i sqrt3 thanks in advance
Evita Ammersingh , 6 Years ago
Grade 12th pass
Integral Calculus> hello, how do i solve : (Z-2i)^4 = -8-8i ...
1 Answers
Aditya Gupta
Last Activity: 6 Years ago
this ques is EXTREMELY easy.
all we need to do is to write RHS in re^itheta form and then apply de moivre rule.theorem.
note that -8-8i sqrt3= 16*e^(i4π/3)
also 16=2^4. let z= 2x
so we have (x – i)^4 = e^(i4π/3)
now x – i= [e^(i4π/3)]1/4 , which will have four distinct values due to moivres rule.
hence x= i + [e^(i4π/3)]1/4
or z= 2i+2[e^(i4π/3)]1/4
note that it is fairly elementary and easy to apply moivres theorem on [e^(i4π/3)]1/4 so you can do that on your own
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