The square root of −3−4i cannot be simplified. The magnitude is the absolute value, or a number's distance from zero. Do it yourself, it's not too tricky, try searching up what magnitude is.
I suppose you could rewrite the square root of the complex number as
i*sqrt(3)+2i*sqrt(i)
Find the square root of the complex number, calculate the magnitude of the number and its square roots
z=−3−4i
3 answers
use De Moivre's Theorem
z = -3 - 4i
magnitude = √(9 + 16)
= 5
argument: tanθ = -4/-3
θ = 233.13 , since -3 - 4i would be in quad III of the Argand plane
z = -3 - 4i
= 5(cos233.13 + i sin233.13)
z^(1/2) = 5^(1/2)(cos116.565 + i sin116.565)
= √5(-.4472... + i(.8944...) )
= -1 + 2i <----- primary root
2nd root:
z^(1/2) = 5^(1/2)(cos(116.565+180) + i sin(116.565+180))
= 1 -2i
Check:
Square each of my answers, you will get -3 -4i
e.g. the last one
(1-2i)^2 = 1 - 4i + 4i^2
= 1 - 4i - 4
= -3 - 4i
z = -3 - 4i
magnitude = √(9 + 16)
= 5
argument: tanθ = -4/-3
θ = 233.13 , since -3 - 4i would be in quad III of the Argand plane
z = -3 - 4i
= 5(cos233.13 + i sin233.13)
z^(1/2) = 5^(1/2)(cos116.565 + i sin116.565)
= √5(-.4472... + i(.8944...) )
= -1 + 2i <----- primary root
2nd root:
z^(1/2) = 5^(1/2)(cos(116.565+180) + i sin(116.565+180))
= 1 -2i
Check:
Square each of my answers, you will get -3 -4i
e.g. the last one
(1-2i)^2 = 1 - 4i + 4i^2
= 1 - 4i - 4
= -3 - 4i
Ah, that's a much more clever way than what I came up with... unfortunately my math isn't advanced enough for that :')