Final answer:
To find the length of the square roots of 3i on the complex plane, we solve for the square root of the complex number 3i, which appears to be ±√(3/2) + ±√(3/2)i. The modulus of this complex number is √3, thus the length of the square roots of 3i is √3 units.
Step-by-step explanation:
To find the length of the square roots of 3i on the complex plane, we need to compute the square roots of a complex number. The complex number 3i lies entirely on the imaginary axis at a distance of 3 from the origin. To find the square roots, we take the principal square root of the complex number, which involves finding a number that, when squared, yields the original complex number.
Let's denote the square root of 3i as z, where z = a + bi, and i is the imaginary unit. Squaring both sides gives:
(a + bi)2 = 3i
a2 - b2 + 2abi = 3i
Equating the imaginary parts gives us 2ab = 3, and since the real part must be 0 (a2 - b2 = 0), we conclude that a = b or a = -b. Substituting into the equation 2ab = 3 and solving for a and b, we get:
a = b = ±√(3/2)
Therefore, the length of the square root of 3i, which is the modulus of z, is equal to:
|z| = √(a2 + b2) = √(3/2 + 3/2) = √3
The square roots of 3i on the complex plane have a length of √3 units.