Find all roots of the equation z^5 = i, i.e. find the five values of i^1/5 and show them on an Argand diagram.
take a look at
http://www.wolframalpha.com/input/?i=solve+z^5+%3D+i
Note that the angles given are
pi/2 + k*2pi/5
To find all the roots of the equation z^5 = i, we can use De Moivre's formula, which states that for any complex number z = r(cosθ + isinθ), its nth root can be found using the formula z^(1/n) = r^(1/n)(cos(θ/n + 2πk/n) + isin(θ/n + 2πk/n)), where k is an integer between 0 and n-1.
In this case, we are looking for the fifth root of i, which can be expressed as i^(1/5). From previous knowledge, we know that i can be written as i = cos(π/2) + isin(π/2). Therefore, r (the magnitude) is 1, and θ (the argument) is π/2.
Substituting these values into the formula, we get:
i^(1/5) = 1^(1/5)(cos(π/2*1/5 + 2πk/5) + isin(π/2*1/5 + 2πk/5))
= cos(π/10 + 2πk/5) + isin(π/10 + 2πk/5)
To find the five values of i^(1/5), we can plug k = 0, 1, 2, 3, 4 into the formula and simplify:
For k = 0: i^(1/5) = cos(π/10) + isin(π/10)
For k = 1: i^(1/5) = cos(3π/10) + isin(3π/10)
For k = 2: i^(1/5) = cos(5π/10) + isin(5π/10) = cos(π/2) + isin(π/2) = i
For k = 3: i^(1/5) = cos(7π/10) + isin(7π/10)
For k = 4: i^(1/5) = cos(9π/10) + isin(9π/10)
Now, let's plot these five points on an Argand diagram. Each point corresponds to one of the values of i^(1/5):
Point 1: cos(π/10) + isin(π/10)
Point 2: cos(3π/10) + isin(3π/10)
Point 3: i (which is equivalent to (1, 0) on the complex plane)
Point 4: cos(7π/10) + isin(7π/10)
Point 5: cos(9π/10) + isin(9π/10)
By plotting these points on the Argand diagram, you will be able to visualize the five roots of the equation z^5 = i.