# 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 $z = -\frac{2ai}{1-a}$