Find the square root of the complex number, calculate the magnitude of the number and its square roots
z=−3−4i
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
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)
Ah, that's a much more clever way than what I came up with... unfortunately my math isn't advanced enough for that :')
To find the square root of a complex number, we can use the square root formula for complex numbers.
Let's find the square root of the complex number z = -3-4i:
Step 1: Write the complex number in the form a + bi, where a is the real part and b is the imaginary part.
For z = -3-4i, a = -3 and b = -4.
Step 2: Find the magnitude (or absolute value) of the complex number using the formula:
|z| = √(a^2 + b^2)
For z = -3-4i, the magnitude is:
|z| = √((-3)^2 + (-4)^2) = √(9 + 16) = √25 = 5.
Step 3: Now, let's find the square root of the complex number.
The square root of a complex number can be found using the formula:
√(z) = ±√((|z| + a)/2) + (sign(b) * √((|z| - a)/2)) * i
For z = -3-4i, the square root can be calculated as follows:
√(z) = ±√((5 - 3)/2) + (-1 * √((5 + 3)/2)) * i
= ±√(2/2) + (-1 * √(8/2)) * i
= ±√1 - √4 * i
= ±(1 - 2i)
Therefore, the square roots of the complex number z = -3-4i are ±(1 - 2i).