Homework Help: Z^4 = 16i

1. Dec 25, 2005

Physics_wiz

Need to find all z such that z^4 = 16i. Rectangular form and no trig functions. Here's what I did:
z^4 = 16e^(i pi/2) = 16e^i(pi/2 + 2npi)

z = 2e^i(pi/8 + npi/2)

First question: Do I add a 2npi before I take the 4th root or do I add it after I take the 4th root to get z = 2e^i(pi/8 + 2npi)? Does it matter?

Second question: After I get an expression for z, which n's do I plug in the equation to find the 4 z's I'm looking for? How do I know that?

2. Dec 25, 2005

marlon

First question : BEFORE
second question : take n : 0,1,2,3

marlon

3. Dec 25, 2005

Unco

G'day, Physics_wiz.

It must be before. You might write it as
Let
$${z_n}^4 = 16e^{i\left(\frac{\pi}{2} + 2n\pi \right)}$$

So
$$z_n = 2e^{i\left(\frac{\frac{\pi}{2} + 2n\pi}{4}\right)} = 2e^{i\left(\frac{\pi}{8} + \frac{n\pi}{2}\right)}$$

The four complex roots will be an angle of $$\frac{2\pi}{4}$$ apart on an Argand diagram.

Any consecutive four; that is how many complex fourth roots we expect of $$z^4$$. eg. n=0, 1, 2, 3; or n=4, 5, 6, 7; etc. The convention is to choose consecutive values for n such that the argument is no greater than pi, though.

Note that you will need the trig functions on your calculator to convert the roots you find in exponential form to rectangular form, so I'm not exactly sure what the question means by "no trig functions".

Last edited: Dec 25, 2005
4. Dec 25, 2005

marlon

No, it must be :
$${z_n}^4 = 16e^{ i \left (\frac{\pi}{2} + 2n\pi \right)}$$
marlon

Last edited: Dec 25, 2005
5. Dec 25, 2005

Unco

Thank you for the correction. Is there a way to preview the Latex output?

6. Dec 25, 2005

marlon

Unco,

Sorry, not right now. Look at this message

regards
marlon