Question

Solve z^3 = 4 sqrt(2*(i-1)). I get nowhere.

Answer 1

Please, is there no one who fancy some quizzy math prob?

Answer 2

@mathmale , could you maybe take a look?

Answer 3

@Lena772 ?

Answer 4

@thomaster

Answer 5

Ummm, consider \((a+bi)^3=4\sqrt{2(i-1)}\)

Answer 6

I'd start by squaring both sides: \[
(a+bi)^3=16(2(i-1)) = -32+32i
\]

Answer 7

\[
(a+bi)^2 =a^2+2abi - b^2
\]

Answer 8

\[
(a+bi)(a^2-b^2+2abi) = (\ldots) +(\ldots)i
\]

Answer 9

When left side is squared it will be (a + bi)^6

Answer 10

how did u stumble upon this ugly question ?

Answer 11

i would try to break this problem in 2 parts
first find

x+iy = 4 sqrt(2*(i-1))
square both sides
and compare real and imaginary parts
to get x and y

then solving z^3 = x+iy

Answer 12

z^6 = 32(-1 + i)
z = { 32(-1 + i) }^1/6
= 2^5/6 * (-1 + i)^1/6
Put (-1 + i) in r(cos(theta) + isin(theta))
Then use DeMoivre's Theorem to raise it to ^1/6.

Answer 13

\[-1 + i = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})\]

Answer 14

now solve.