SOLUTION: (i need help with part B of this question) The complex number v has a modulus 1 and argument pi/6 and the complex number w has a modulus 2 and argument -2pi/3 a)Express wv and

Question 816172: (i need help with part B of this question)
The complex number v has a modulus 1 and argument pi/6 and the complex number w has a modulus 2 and argument -2pi/3
a)Express wv and iv in mod-arg form where the argument is between -pi and pi.
Solution : wv = 2cis(-pi/2) iv = cis(2pi/3)
b)Show that v is a solution of the equation z^4=iz, and state the non-zero roots (z1,z2 and z3) of this equation.
I proved that v is a solution of z^4=iz by substituting v into the equation and getting the value of iv, which is cis(2pi/3)
I do not know how to find the non zero roots of the equation.
Thank you!
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

Subtracting iz:

Factor out z:

From this we can see that z = 0 will be a root. The other (non-zero) roots will be solutions to . In other words the other (non-zero) roots are the cube roots of i:

In mod-arg form:

To find these cube roots we will use DeMoivre's (spelling?) Theorem which tells us a power of a complex number in mod-arg form can be found by raising the mod to the power and multiplying the arg by the power:

which simplifies to:

This is the root, v, that you had already found. To find the other roots we just use co-terminal args:

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