Find the square root of 4-3i?

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Answer 1 · 2018-08-07T17:35:30

#z = pm (3 - i)/2 sqrt 2#

#z^2 = 4 - 3i = (x + yi)^2 = x^2 - y^2 + 2xyi#

#x^2 - y^2 = 4#
#2xy = -3 => y = -3/(2x)#

#x^2 - 9/(4x^2) = 4, x^2 = a#

#4a^2 - 16a - 9 = 0#

#Delta = 256 + 4 * 4 * 9 = 400#

#x^2 = (16 pm 20)/8 > 0#

#x = pm 3/sqrt2 = pm 3/2 sqrt 2#

#y = -pm3/2 * sqrt 2/3 = - pm 1/2 sqrt 2#

Answer 2 · 2018-08-08T14:12:05

#4-3i#

#=1/2(8-2.3*i)#

#=1/2(3^2+i^2-2.3*i)#

#=1/2(3-i)^2#

Square root of #4-3i#

#pm1/sqrt2(3-i)#