#1
hey guys,
I've come to the pit a couple of times for maths or science help and you haven't failed me yet, but now i seriously need help for this question,

The complex number v has modulus 1 and argument pi/6
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.

Thanks in advance
Quote by yankovicfan3125
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#5
I think that for this type of question, use De Moivre's theorm:


theta is the argument, and r is the modulus

(r*cis(theta))^n=(r^n)*cis(n*theta)

Also remember: r*cis(alpha)*a*cis(beta)=r*a*cis(alpha+beta)

use the first fact to get the 3 roots, and the second to show that the equation is satisfied.

Hope this works (been a while since I last did complex numbers).
"Feelin' stupid? I know I am!"
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#7
OK, I've found values for a and b, I'll have a crack at solving the rest, but I've got coursework of my own, so don't wait up on me either. Or don't count on me.
#8
The answer is C
You who build these altars now

To sacrifice these children
You must not do it anymore
#9
z^4 = iz
z^3 = i
|z^3| = 1
arg(z^3) = pi/2

z^3 = e^i(pi/2) = e^i(pi/2 + 2k*pi), where k is a constant
z = e^i(1/3)(pi/2 + 2k*pi)
for (1/3)(pi/2 + 2k*pi) to be in the range of -pi to pi, k = 0,1,-1

when k = 0, z = e^i(pi/6) = v
when k = 1, z = e^i(5/6 pi)
when k =-1, z = e^i(-pi/2)