How to Solve Equations with Complex Numbers

[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 e^iπ/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.

 

[SteamKing]

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 e^iπ/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.

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.

 

[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 $\sqrt[4]{i}$?

 

[cdummie]

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 $\sqrt[4]{i}$?

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.

 

[SteamKing]

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.

It's still not clear how calculating i^1/4 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 $f(z) = 0$?

 

[cdummie]

It's still not clear how calculating i^1/4 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 $f(z) = 0$?

Well, i thought, if i find the exact value on the rhs (that is e^iπ/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?

 

[SteamKing]

Well, i thought, if i find the exact value on the rhs (that is e^iπ/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?

Yes.

 

[HallsofIvy]

It's still not clear how calculating i^1/4 gets you any closer to finding the value of z which solves the original equation.

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 $z= -\frac{ai}{1- a}$.

 

[LAZYANGEL]

Use these values for the right hand side:

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

 

[jbriggs444]

$z= az- 2ai$ so that $z- az= (1- a)z= -ai$ and then $z= -\frac{ai}{1- a}$.

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