Demoivre's Theorom and nth roots

[mynamesmurph]

Find the complex 5th roots of z=-1 - i and write in rectangular form.
(
I haven't had any trouble with these except for this one. Anyway, here's what I got.

Θ=5π/4 r=√2

so I set it up like this


For the first root, where k=0

2^(1/10)*{cos[(5π/4+2π(0)/5]+sin[(5π/4+2π(0)/5]}

2^(1/10)*cos(π/4)+i*sin(π/4)

2^(1/10)*[(√2)/2+i(√2)/2]


I get (2^(3/5))/2 + i(2^(3/5))/2

But the answer is in the book is (2^(1/10))/2 + i(2^(1/10)/2

 

[Ishuda]

Find the complex 5th roots of z=-1 - i and write in rectangular form.
(
I haven't had any trouble with these except for this one. Anyway, here's what I got.

Θ=5π/4 r=√2

so I set it up like this


For the first root, where k=0

2^(1/10)*{cos[(5π/4+2π(0)/5]+sin[(5π/4+2π(0)/5]}

2^(1/10)*cos(π/4)+i*sin(π/4)

2^(1/10)*[(√2)/2+i(√2)/2]


I get (2^(3/5))/2 + i(2^(3/5))/2

But the answer is in the book is (2^(1/10))/2 + i(2^(1/10)/2

Well I have been known to be wrong but I would take your answer over the books if that's what they had [after, of course, you correct your typo of 3/5 to 2/5 (unless I'm wrong again)].

 

[mynamesmurph]

Isn't it 2^(3/5)

2^(1/10)*2(1/2)=2^(3/5) right?

 

[Ishuda]

Isn't it 2^(3/5)

2^(1/10)*2(1/2)=2^(3/5) right?

Of course you're correct - I had encompassed the 2 in the denominator and just wrote the complete thing as 2^-2/5 and then dropped the minus sign.