Complex roots of $z^{2018} =2018^{2018} +i$ in complex plane

Question

Number of solution of $$z^{2018} =2018^{2018} +i$$ where $i=\sqrt{-1~}$. How are the roots positioned in $4$ quadrants ?

Answer 1

Note,

$$z =(2018^{2018} +i )^{\frac 1{2018}}= re^{i\theta}$$

where $$r = (2018^{2\cdot 2018} + 1)^{\frac 1{2\cdot2018}}\approx 2018$$ $$\theta= \tan^{-1}\left(\frac1{2018^{2018}}\right)+\frac{2\pi n}{2018}\approx \frac{2\pi n}{2018}$$

with $n =0, 1,2,\>...\>2017$. Thus,

$$z = 2018 e^{\frac{i2\pi n}{2018}}$$

representing 2018 distinctive roots evenly spaced around a circle of radius 2018 in the complex plane.