find the trigonometric form of the complex number - (square root) 3 + i
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
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.
To find the trigonometric form of a complex number, we can use the formula:
z = r * (cosθ + isinθ)
where z is the complex number, r is the magnitude or modulus, and θ is the argument or phase angle.
To begin, let's find the magnitude or modulus (r) of the complex number -√3 + i using the formula:
|r| = √(Re^2 + Im^2)
where Re is the real part and Im is the imaginary part.
In this case, Re = -√3 and Im = 1, so we have:
|r| = √((-√3)^2 + 1^2)
|r| = √(3 + 1)
|r| = √4
|r| = 2
Now, let's find the argument or phase angle (θ) using the formula:
θ = tan^(-1)(Im/Re)
In this case, Re = -√3 and Im = 1, so we have:
θ = tan^(-1)(1/(-√3))
θ = tan^(-1)(-1/√3)
θ ≈ -30°
Since the real part is negative, the angle is in the third quadrant.
Now, let's substitute the values we found into the formula for the trigonometric form:
z = 2 * (cos(-30°) + isin(-30°))
Using the trigonometric identities, we have:
z = 2 * (√3/2 - i/2)
z = √3 - i
Therefore, the trigonometric form of the complex number -√3 + i is √3 - i.