Question

Find all complex solution to the equation z^6 = 64.

Answer 1

Any help appreciated!

Answer 2

\[z = \sqrt[6]{64}\]

Answer 3

So z = 2. There two possibilities which is 2 ad -2. -2i would be the complex root if 64 was negative.

Answer 4

So is the only possible complex root of the equation -2i, or would there be no complex root with the equation in its current configuration?

Answer 5

Yes. Complex roots only arise when the value is negative in the square root sign. So technically this equation only has real roots. If this equation was \[z =\sqrt[6]{-64}\]
then you would have -2i and 2i as your roots. Since 64 is positive you would have 2 and -2.

Answer 6

It is not enough to find all the complex roots out. You have to go some more steps.

Answer 7

Or, I am making it so complicated. :)

Answer 8

That is about it. No extra work is needed.

Answer 9

Like this, if z=2, then \(z=|2|e^{i*0}\)
Hence there exists a \(w\) as a root of z such that \(z^6 =64\)
Let \(w=|w|e^{i\tau}\)
\(w^6=|w|^6 e^{6i\tau}\)
Hence \(|w| =2\)
Now, imaginary part
\(6i\tau =i*0 +2i kpi \)
hence \(\tau = k\pi/3, 0\leq k\leq 5\)
for k =0, you have \(w=2\)
for k =1, you have \(w=2e^{i\pi/3}=2(cos(\pi/3)+isin(\pi/3)=1+i\sqrt 3\)
Now, check up \((1+i\sqrt 3)^6=64\)
Hence, one of value of z is \(1+i\sqrt 3\)
Do the same with k =2,3,4,5 to find all of complex roots of z

Answer 10

This is the proof of \((1+i\sqrt 3)^6=64\)
https://www.wolframalpha.com/input/?i=(1%2Bisqrt(3))%5E6

Answer 11

@snowsurf I think I complicated the stuff. hehehe....

Answer 12

Haha.

That is a cool result. I would never had thought of that.

Answer 13

:) it is from my complex analysis course.

Answer 14

So you applied Euler's formula

Answer 15

\[z^6=64=r \left( \cos \theta+\iota \sin \theta \right)\]
\[r \cos \theta=64,r \sin \theta=0,\]
square and add
\[r^2\left( \cos ^2 \theta +\sin ^2\theta \right)=64^2+0^2,r=64\]
Hence \[\cos \theta=1=\cos 0=\cos( 0+2 n \pi),\theta=2n \pi\]
where n is an integer.
\[z^6=64(\cos 2 n \pi+\iota \sin 2n \pi)=64e ^{2n \pi}\]
\[z=64^{\frac{ 1 }{ 6 }}e ^{\frac{ 2n \pi }{ 6 }}=2e ^{\frac{ n \pi }{ 3 }}\]
put n=0,1,2,3,4,5 and get the solutions.

Answer 16

That simultaneously makes more sense and is more complicated. Haha I like it

Answer 17

correction
it is \[z^6=64 e ^{\iota 2n \pi}\]
then \[z=2e^\iota \frac{ 2n \pi }{ 3 }\]