How do i solve this problem?
Find the polar form of the complex number where the argument satisfies 0≤θ≤2π.
0+3i
the real part is zero, so it's a pure imaginary: on the y-axis
0+3i = (3,π/2)
of course, it could also be (-3,3π/2) ...
Doh! I’m an idiot.
To find the polar form of a complex number, you need to determine its magnitude (r) and argument (θ).
For the complex number 0 + 3i, we can see that the real part is 0 and the imaginary part is 3.
To find the magnitude (r), we use the formula:
r = √(real^2 + imaginary^2)
In this case, r = √(0^2 + 3^2) = √9 = 3.
Next, to find the argument (θ), we use the formula:
θ = arctan(imaginary/real)
In this case, since the real part is 0, we have θ = arctan(3/0). However, this is undefined since the tangent function approaches infinity when the input is 90 degrees. So, we will find the argument using a different approach.
Since the imaginary part is positive (+3i), the argument lies in the positive y-axis. It forms a 90-degree angle with the positive x-axis.
Therefore, the argument (θ) is π/2 (90 degrees).
The polar form of the complex number 0 + 3i is therefore:
3(cos(π/2) + i sin(π/2))
This can also be written as:
3eiπ/2
So, the polar form of the complex number 0 + 3i, where the argument satisfies 0 ≤ θ ≤ 2π, is 3eiπ/2.