# complex no.

• May 1st 2007, 11:19 AM
complex no.
Hey All,

How would you find the following complex number problem:

find the four roots of the equation z^4= j3, expressing yuor answer in the form re^{jθ}, where r and θ are real. sketch their positions in an Argand diagram.

Thanks
• May 1st 2007, 11:21 AM
ThePerfectHacker
Quote:

Hey All,

How would you find the following complex number problem:

find the four roots of the equation z^4= j3, expressing yuor answer in the form re^{jθ}, where r and θ are real. sketch their positions in an Argand diagram.

Thanks

But j3 you mean i^3 ?
• May 1st 2007, 11:25 AM
well the question i copied it from is j3. unless their was a typo?

buy don't complex numbers have the form a +jb
• May 1st 2007, 12:25 PM
topsquark
Quote:

well the question i copied it from is j3. unless their was a typo?

buy don't complex numbers have the form a +jb

Typically most Mathematicians (and Physicists) use i^2 = -1. Engineers typically use j^2 = -1. Though, of course, there are exceptions to that rule.

-Dan
• May 1st 2007, 01:03 PM
yes this is from an engineering book where j^2= -1

find the four roots of the equation z^4= -16, expressing your answer in the form a+jb, where a and b are real. sketch their positions in an Argand diagram.
• May 1st 2007, 03:10 PM
CaptainBlack
Quote:

Hey All,

How would you find the following complex number problem:

find the four roots of the equation z^4= j3, expressing yuor answer in the form re^{jθ}, where r and θ are real. sketch their positions in an Argand diagram.

Thanks

First you need to know or be able to derive:

j = e^{pi j/2 + 2 pi j n}, n=0, +/-1, ...

this is because: e^{pi j/2} = cos(pi/2) + j sin(pi/2) = j.

Now z^4 = j^3 = e^{3 pi j/2 + 6 pi j n}, and so:

z = e^{[3 pi j/2 + 6 pi j n]/4} = e^{3 pi j/8 + 3/2 pi j n},

Putting n =0, 1, 2, 3 should give four distinct values for z, and any
other value will again give one of these.

RonL
• May 2nd 2007, 05:26 AM