If #z^3 = -11-2i# then what is #z# ?

1 Answer
George C.
Nov 18, 2017

Gaussian integer solution:

#z = 1+2i#

Other solutions:

#z = -1/2(1+2sqrt(3))+1/2(-2+sqrt(3))i#

#z = -1/2(-1+2sqrt(3))+1/2(-2-sqrt(3))i#

Explanation:

Taking the modulus of a complex number preserves multiplication. That is:

#abs(z_1 z_2) = abs(z_1) * abs(z_2)#

Note that:

#abs(-11-2i) = sqrt(11^2+2^2) = sqrt(121+4) = sqrt(125) = sqrt(5^3)#

So given that #abs(z^3) = sqrt(5^3)#, we must have #abs(z) = sqrt(5)#

That leaves us with #8# possible Gaussian integers:

#+-1+-2i" "# or #" "+-2+-i#

Note that #-11-2i# is in Q3, just below the real axis. So one of the cube roots will be in Q1, just above the #pi/3# line.

So try #(1+2i)^3#:

#(1+2i)^3 = 1+3(2i)+3(2i)^2+(2i)^3#

#color(white)((1+2i)^3) = 1+6i-12-8i#

#color(white)((1+2i)^3) = -11-2i" "# as required.

The other two cube roots will have irrational real or imaginary coefficients. We can find them by multiplying by: #omega# or #omega^2# where:

#omega = -1/2+sqrt(3)/2i#

is the primitive complex cube root of #1#.

#omega(1+2i) = 1/2(-1+sqrt(3)i)(1+2i)#

#color(white)(omega(1+2i)) = 1/2(-1-2i+sqrt(3)i-2sqrt(3))#

#color(white)(omega(1+2i)) = 1/2(-(1+2sqrt(3))+(-2+sqrt(3))i)#

#color(white)(omega(1+2i)) = -1/2(1+2sqrt(3))+1/2(-2+sqrt(3))i#

#omega^2(1+2i) = 1/2(-1-sqrt(3)i)(1+2i)#

#color(white)(omega^2(1+2i)) = 1/2(-1-2i-sqrt(3)i+2sqrt(3))#

#color(white)(omega^2(1+2i)) = 1/2((-1+2sqrt(3))+(-2-sqrt(3))i)#

#color(white)(omega^2(1+2i)) = 1/2(-1+2sqrt(3))+1/2(-2-sqrt(3))i#

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