Describe the solutions of the equation in terms of roots of unity?

Question

I want to find the solutions of the equation $$\left[z- \left( 4+\frac{1}{2}i\right)\right]^k = 1 $$

in terms of roots of unity.

When I try to solve this, I get \begin{align*}z - 4 - \dfrac i2 &= 1\\ z-\dfrac{i}{2}&=5\\ \dfrac{2z-i}2 &= 5\\ z&= 5 + \dfrac i2\end{align*}

Is this the right approach?

I want to do the same for $$\left[z-\left(4+\frac{1}{2}i\right)\right]^k = 2$$

as well.

Answer 1

HINT: You should find (or only name?) $w_0$, $w_1$, ... $w_{k-1}$, which are $k$-th roots of 1 (2, respectively) and compare them one by one with $z-4+\frac12i$.

Answer 2

Let $w = z - 4 + \dfrac{i}{2} \to w^k = 1 \to w = e^{\left(\dfrac{2\pi in}{k}\right)}, n = 0,1,\cdots ,(k-1) \to z = e^{\left(\dfrac{2\pi in}{k}\right)} + 4 - \dfrac{i}{2}$,

For b), $w^k = 2 = \left(\sqrt[k]{2}\right)^k \to \left(\dfrac{w}{\sqrt[k]{2}}\right)^k = 1 \to \dfrac{w}{\sqrt[k]{2}} = e^{\left(\dfrac{2\pi in}{k}\right)} \to w = \sqrt[k]{2}\cdot e^{\left(\dfrac{2\pi in}{k}\right)} \to z = \sqrt[k]{2}\cdot e^{\left(\dfrac{2\pi in}{k}\right)} + 4 - \dfrac{i}{2}, n = 0,1,\cdots ,(k-1)$.

Answer 3

Things you need to know (Hints):

$$\left(z-(4+\frac{1}{2}i)\right)^k=\color{Crimson}{1}$$

$$ \color{crimson}{\cos 2\pi n+ i \sin 2\pi n =1}$$

$$\text{Where n is an Integer}$$

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Also you should know De Moivre's Theorem:

$$( \cos \theta+ i \sin \theta)^b= \cos b\theta+ i \sin b\theta$$

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Also when it comes to second part:

$$\left(z-(4+\frac{1}{2}i)\right)^k=2$$

$$\left(z-(4+\frac{1}{2}i)\right)^k=2(\color{Crimson}{1})$$