Roots of complex numbers
Roots of complex numbersShowing the nth roots of complex numbers are evenly spacedConsider solving (a+ib)1nz=(a+ib)1nLet z=rcisθrcisθ=a2+b2cis(α+2πk)1n Where k=0,1,2,..(n-1) and tanα=abBy De Moivre' s theoremrcisθ=na2+b2 cisα+2πknr=na2+b2, θ=αn, αn+2πn, αn+4πn,…, αn+2(n-1)πnThe nth roots of a complex number lie on a circle with radius na2+b2 and are evenly spaced by equal length arcs which subtend angles of 2πn at the origin.Note: This could be modelled using a numerical example.Finding the nth root of complex numbersSolve 2i12z=2i12Let z=rcisθrcisθ=2cisπ212 By De Moivre' s theoremrcisθ=2cisπ4There will be 2 roots which are evenly spread over 2π (one revolution).They will be 2π2=π radians apart.r=2 and θ=π4, 5π4Polar form: z=2cisπ4 and 2cis5π4Cartesian form:z=2cosπ4+isinπ4=222+i22=1+iz=2cos5π4+isin5π4=2-22+i-22=-1-i Solve 3-i13z=3-i13Let z=rcisθrcisθ=2cis11π613By De Moivre' s theoremrcisθ=32cis11π18There will be 3 roots which are evenly spread over 2π (one revolution).They will be 2π3 or 12π18 radians apart.r=32 and θ=11π18, 23π18,35π18Polar form: z=32cis11π18 , 32cis23π18 and 32cis35π18 ................
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