#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

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

#2

wolfram alpha, go there

#4

I do Advanced Higher Maths, I think I could help.

Too bad I'm busy having breakfast though.

Too bad I'm busy having breakfast though.

#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).

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).

#6

Too bad I'm busy having breakfast though.

#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

#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)

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)