# complex no. again

• May 22nd 2008, 08:28 AM
afeasfaerw23231233
complex no. again
p306
in my textbook it writes:
given: $\displaystyle z^4 =1+i$
$\displaystyle z = [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4 )]^ {\frac 1 4}$
where does the $\displaystyle \pm$ go?
i don't understand why it shouldn't be $\displaystyle z = \pm [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4)]^ {\frac 1 4}$
• May 22nd 2008, 08:30 AM
colby2152
Quote:

Originally Posted by afeasfaerw23231233
p306
in my textbook it writes:
given: $\displaystyle z^4 =1+i$
$\displaystyle z = [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4 )]^ {\frac 1 4}$
where does the $\displaystyle \pm$ go?
i don't understand why it shouldn't be $\displaystyle z = \pm [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4)]^ {\frac 1 4}$

Is there a constraint?
• May 22nd 2008, 08:34 AM
bobak
Quote:

Originally Posted by afeasfaerw23231233
p306
in my textbook it writes:
given: $\displaystyle z^4 =1+i$
$\displaystyle z = [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4 )]^ {\frac 1 4}$
where does the $\displaystyle \pm$ go?
i don't understand why it shouldn't be $\displaystyle z = \pm [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4)]^ {\frac 1 4}$

A little note. If you have $\displaystyle x^2 = y^2$ the solution are given by $\displaystyle x = \pm y$. The general solution $\displaystyle x^4 = y^4$ is not just $\displaystyle x = \pm y$ you also have $\displaystyle x = \pm i y$ as a solution as well.

Bobak
• May 22nd 2008, 11:33 AM
afeasfaerw23231233
no. it doesn't have a constraint. it's an example in my textbook:
e.g.3
Find the fourth roots of 1+i
solution:
let $\displaystyle z^4 =1+i$
$\displaystyle z = [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4 )]^ {\frac 1 4}$
yakyakyak.... ,etc
• May 22nd 2008, 04:01 PM
mr fantastic
Quote:

Originally Posted by afeasfaerw23231233
p306
in my textbook it writes:
given: $\displaystyle z^4 =1+i$
$\displaystyle z = [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4 )]^ {\frac 1 4}$
where does the $\displaystyle \pm$ go?
i don't understand why it shouldn't be $\displaystyle z = \pm [\sqrt 2 (cos \frac \pi 4 + i sin\frac \pi 4)]^ {\frac 1 4}$

The book is wrong.

It should say:

$\displaystyle z = \left( \sqrt 2 \left[ \cos \left( \frac \pi 4 + 2 n \pi \right) + i \sin \left( \frac \pi 4 + 2 n \pi \right) \right] \right)^{\frac 1 4}$

where n is an integer.

When n = 0 you get one of the real fourth roots. This is the one given in the book.

There is obviously another real fourth root, as you rightly point out. It's given by n = 2.

Then there are two other fourth roots, both non-real, given by n = -1 and n = 1. These are the ones that bobak makes mention of.

Four fourth roots in total, as expected.

Other values of n just give you the same four roots again.