find the trigonometric form of the complex number - (square root) 3 + i

Answer 1

let z = √3 + i
magnitude = √(√3^2 + 1) = 2
angle Ø:
tanØ = 1/√3
Ø = π/6 or 30°

so z = 2(cos30° + i sin30°) or 2cis 30°
or 2 cis π/6

√z = z^(1/2)
= √2( cos 15° + i sin15°) or √2(cos π/12 + i sin π/12)
or √2 cis π/12

for the nth root of a complex number, there will be n such solutions, so for the square root we need 2 answers
2π/2 = π OR 360°/2 = 180°
so by adding π radians or 180° to the appropriate form we get our second solution.

so √z = √2cos15° + √2sin15° or √2cos195° + √2sin195°
or
√z = √2cos π/12 + √2sin π/12 or √2cos 13π/12 + i sin 13π/12

in short form:
√z = √2cis π/12 , √2cis 13π/12

Answer 2

just noticed the - sign in front of √3
so our z = -√3 + i

no big deal:
the magnitude stays the same,
however, the angle would now be in quad II
so Ø = 150° or 5π/6

making our
√z = √2(cos 75° + i sin75°) OR √2( cos 5π/12 + i sin 5π/12)

I will let you make these and the other changes for the second square root.