Equation with complex numbers

1. Nov 12, 2015

cdummie

I have to solve the following equation:

z4=i*(z-2i)4

Now, i tried to move everything but i (imaginary number) to the left side and then find the 4-th root of i, when i did that, i had four solutions, with one of them being eiπ/8. But i don't know what to do with the left side, since i get way to complicated expression. So there must be less complicated way to solve this.

2. Nov 12, 2015

SteamKing

Staff Emeritus
It's not clear what you did here. Can you share the other steps you made to get everything but i to one side of the equation?

After doing the algebra, it doesn't seem like you would wind up with a simple equation like z4 - i = 0, which has as its solution of the fourth roots of i.

3. Nov 12, 2015

HallsofIvy

If I understand you, you divided both sides by $z- 2i$ to get $\frac{z^4}{(z- 2i)^4}= \sqrt[4]{i}$. The right side has, of course, four different values. What did you get for the four values of $\sqrt[4]{i}$?

4. Nov 12, 2015

cdummie

Actually, i divided both sides by (z-2i)4 but i assume that you meant that.

Anyway, I got the following:

i= 0+i = cos(π/2) + isin(π/2)

then, fourth root of i is:

cos[(π/2 + 2kπ)/4) + isin[(π/2 + 2kπ)/4)

so i have the following 4 solutions:

i0=cos(π/8) + isin(π/8)
i1=cos(5π/8) + isin(5π/8)
i2=cos(9π/8) + isin(9π/8)
i3=cos(13π/8) + isin(13π/8)

but i still don't know what can i do next?

@SteamKing I did exactly what HallsofIvy said.

5. Nov 12, 2015

SteamKing

Staff Emeritus
It's still not clear how calculating i1/4 gets you any closer to finding the value of z which solves the original equation.

After all,

$\frac{z^4}{(z-2i)^4}=i^4$

but you are looking for the value of z which satisfies the equation $z^4 = i (z - 2i)^4$.

Why didn't you expand the RHS, collect all the resulting terms on the LHS, and then solve the resulting polynomial $f(z) = 0$?

6. Nov 12, 2015

cdummie

Well, i thought, if i find the exact value on the rhs (that is eiπ/8 and three other values) and if i have $\frac{z}{z-2i}$ on the lhs then if i write a+ib instead of z (which means that z=a +ib) i could get the solution, but it turns out it isn't working. By expansion, do you mean using binomial theorem?

7. Nov 12, 2015

SteamKing

Staff Emeritus
Yes.

8. Nov 12, 2015

HallsofIvy

Yes, it does. Once you have arrived at $\frac{z}{z- 2i}= a$ (where a is one of the four fourth roots of i) it follows that $z= az- 2ai$ so that $z- az= (1- a)z= -ai$ and then $z= -\frac{ai}{1- a}$.

9. Nov 12, 2015

LAZYANGEL

Use these values for the right hand side:

$i^{\frac{1}{4}}=\left ( e^{i \frac{\pi}{2}}\right )^{\frac{1}{4}}=e^{i \frac{\pi}{8}}=\cos{\left (\frac{\pi}{8}\right)}+i \sin{\left (\frac{\pi}{8} \right )}$

10. Nov 12, 2015

jbriggs444

I think you dropped a factor of 2 in there. $z = az - 2ai$ leads to $z = -\frac{2ai}{1-a}$