Find all complex numbers $z$ satisfying the equation

Question

I need some help on this question. How do I approach this question?

Find all complex numbers $z$ satisfying the equation

$$ (2z - 1)^4 = -16. $$

Should I remove the power of $4$ of $(2z-1)$ and also do the same for $-16$?

Answer 1

HINT: How many solutions does $x^4+16=0$ have?

$x_{1,2,3,4}= \sqrt{2}(\pm 1 \pm i)$, substitute $x$ with $2z-1$, solve for $z$ and you're done.

Answer 2

$(2z-1)^4=2^4e^{(2n+1)\pi i}$ where $n$ is any integer.

So, $$2z-1=2e^{\frac{(2n+1)\pi i}4}=z_n(say)$$ where $n$ can assume any $4$ in-congruent values $\pmod 4,$ the simplest set of values being $0,1,2,3.$

There will be $4$ roots as the given equation is quartic/biquadratic.

$$z_{2+r}=2e^{\frac{\{2(2+r)+1\}\pi i}4}=2e^{\frac{(2r+1)\pi i }4}e^{\pi i}=-z_r--->(1)$$ as $e^{i\pi}=-1$ (using Euler's Identity)

Putting $n=0,z_0=2e^{\frac{\pi i}4}=2\frac{1+i}{\sqrt 2}=\sqrt 2(1+i)$ (Using Euler's Identity)

Putting $n=1,z_1=2e^{\frac{3\pi i}4}=2\frac{1-i}{\sqrt 2}=\sqrt 2(1-i)$

Using $(1), z_2=-z_0,z_3=-z_1$

Answer 3

The answer to this problem lies in roots of a polynomial. From an ocular inspection we know we will have complex roots and they always Come in pairs! The power of our equation is 4 so we know will have a pair of Complex conjugates. I believe it is called the Fundamental Theorem. Addendum A key observation is how to represent -1 in its general term using Euler's identity and the concept of odd powers in EF.