How many roots does the equation $z^2018=2018^2018+i$ have?
How many roots does the equation $z^2018=2018^2018+i$ have?
Consider the equation
$$ z^2018=2018^2018+i$$ where $i=sqrt-1$.
How many complex solutions as well as real solutions does this equation have?
My attempt:
I took the polar form as the equation has very difficult to handle when using $z=x+iy$.
So I set $z=re^iθ$, which yields
$$ (re^iθ)^2018=2018^2018+e^ifracpi2$$
After this I was not able to handle it.
$begingroup$
There are 2018 complex solutions, no real solutions.
$endgroup$
– peterh
Sep 8 '18 at 20:58
$begingroup$
@peterh yes you are right.
$endgroup$
– jayant98
Sep 8 '18 at 21:01
3 Answers
3
hint
If $$z=re^itheta$$ then
$$z^n=r^n(cos(ntheta)+isin(ntheta))$$
the real part gives
$$r^2018cos(2018theta)=2018^2018$$
and the imaginary
$$r^2018sin(2018theta)=1$$
$begingroup$
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
$endgroup$
– jayant98
Sep 8 '18 at 20:54
$begingroup$
I know the result by the way. :/
$endgroup$
– jayant98
Sep 8 '18 at 20:54
$begingroup$
@jayant98 Is it helpful now.
$endgroup$
– hamam_Abdallah
Sep 8 '18 at 20:58
$begingroup$
Thank you. Yes, it is helpful. Sorry, you have to do it again.
$endgroup$
– jayant98
Sep 8 '18 at 20:59
HINT
We have that
$$r=|2018^2018 + i|=sqrt2018^4036+1$$
$$theta =Arg(2018^2018 + i)=arctan left(frac12018^2018right)$$
then
$$2018^2018 + i=sqrt2018^4036+1,e^itheta$$
now use $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
$begingroup$
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
$endgroup$
– jayant98
Sep 8 '18 at 20:57
$begingroup$
@jayant98 You are welcome! Bye
$endgroup$
– gimusi
Sep 8 '18 at 20:58
There are several red herrings in the question and, as the problem is stated, you don't need to describe the solutions.
Your problem can be generalized to $z^n=a+i$, where $n$ is a positive integer and $a$ is real. Clearly, $z$ cannot be real, nor can $a+i$ be zero.
Thus the solutions are all complex (not real) and they are the $n$-th roots of $a+i$. There are $n$ of them.
Thanks for contributing an answer to Mathematics Stack Exchange!
But avoid …
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Required, but never shown
Required, but never shown
By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.
$begingroup$
Sorry, just started to learn the latex use.
$endgroup$
– jayant98
Sep 8 '18 at 20:42