Question

please help what is the fourth root of 8 in the form x + iy?

Answer 1

will the calculations differ from -8 /

Answer 2

Yes because 4 is an even number

Answer 3

I think we should put (x+iy)^4=8 and find x and y

Answer 4

I think this is wrong ?
z⁴ = -8
z⁴ + 8 = 0
(z²)² – (2i√2)² = 0 →→→ Because i²=-1
(z² + 2i√2)(z² – 2i√2) = 0 →→→ a²–b²=(a–b)(a+b)
[z² – (1 – 2i – 1)(√2)][z² – (1 + 2i – 1)(√2)] = 0
[z² – (1² – 2i + i²)(√2)][z² – (1² + 2i – i²)(√2)] = 0
[z² – (1 – i)²(∜2)²][z² – (1 + i)²(∜2)²] = 0 →→→ a²±2ab+b²=(a±b)²
[z – (1 – i)(∜2)][z + (1 – i)(∜2)][z – (1 + i)(∜2)][z + (1 + i)(∜2)] = 0
→→→ a²–b²=(a–b)(a+b)

Answer 5

This is actually correct with a little typo mistake on 6th line [z² – (1² – 2i + i²)(√2)][z² – (1² + 2i + i²)(√2)] = 0 (+ instead of -)
But this is for the 4th root of -8 and we're looking for the 4th root of 8.

Answer 6

Let me think about it for a moment, I'll let you know if I have an answer !

Answer 7

http://www.wolframalpha.com/input/?i=%288+%2B+0i%29%5E%281%2F4%29

Answer 8

Note how the plot converted the 8 to 2^3.

Answer 9

Ok I was able to do it manually here's how :
z⁴ = 8
z⁴ - 8 = 0
(z²)² – (2√2)² = 0
(z² + 2√2)(z² – 2√2) = 0 →→→ a²–b²=(a–b)(a+b)
[z² – (i√2∜2)²][z² – (√2∜2)²] = 0
[z – (i√2∜2)][z + (i√2∜2)][z – (√2∜2)][z + (√2∜2)] = 0 →→→ a²–b²=(a–b)(a+b)

Answer 10

that makes sense but last question.... if I have to get the value of z1 = then how will I get it

Answer 11

Basic equation solving : [z – (i√2∜2)][z + (i√2∜2)][z – (√2∜2)][z + (√2∜2)] = 0
So you have 4 solutions :

z1 = i√2∜2
z2= -i√2∜2
z3= √2∜2
z4 = -√2∜2

Answer 12

thanks that was a stupid question. thanks for your help.

Answer 13

Anytime :)