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How many roots does the equation $z^2018=2018^2018+i$ have?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
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.
complex-numbers roots
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
asked Sep 8 at 20:38
jayant98
579
add a comment |Â
up vote
2
down vote
favorite
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.
complex-numbers roots
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
asked Sep 8 at 20:38
jayant98
579
Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
@peterh yes you are right.
– jayant98
Sep 8 at 21:01
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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.
complex-numbers roots
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
asked Sep 8 at 20:38
jayant98
579
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.
complex-numbers roots
complex-numbers roots
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
asked Sep 8 at 20:38
jayant98
579
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
asked Sep 8 at 20:38
jayant98
579
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
edited Sep 8 at 23:53
AccidentalFourierTransform
1,333627
1,333627
asked Sep 8 at 20:38
jayant98
579
asked Sep 8 at 20:38
jayant98
579
asked Sep 8 at 20:38
jayant98
579
579
Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
@peterh yes you are right.
– jayant98
Sep 8 at 21:01
add a comment |Â
Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
@peterh yes you are right.
– jayant98
Sep 8 at 21:01
Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
@peterh yes you are right.
– jayant98
Sep 8 at 21:01
@peterh yes you are right.
– jayant98
Sep 8 at 21:01
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
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
$$r^2018sin(2018theta)=1$$
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
add a comment |Â
up vote
2
down vote
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 $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
answered Sep 8 at 20:43
gimusi
71.4k73786
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
add a comment |Â
up vote
0
down vote
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.
answered Sep 8 at 21:03
egreg
166k1180187
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
$$r^2018sin(2018theta)=1$$
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
add a comment |Â
up vote
2
down vote
accepted
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
$$r^2018sin(2018theta)=1$$
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
$$r^2018sin(2018theta)=1$$
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
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
$$r^2018sin(2018theta)=1$$
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
edited Sep 8 at 20:57
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
answered Sep 8 at 20:45
Salahamam_ Fatima
33.9k21230
33.9k21230
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
add a comment |Â
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
Sorry, but unable to utilise your hint in the problem. I am not able to understand your hint properly.
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
I know the result by the way. :/
– jayant98
Sep 8 at 20:54
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
@jayant98 Is it helpful now.
– Salahamam_ Fatima
Sep 8 at 20:58
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
Thank you. Yes, it is helpful. Sorry, you have to do it again.
– jayant98
Sep 8 at 20:59
add a comment |Â
up vote
2
down vote
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 $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
answered Sep 8 at 20:43
gimusi
71.4k73786
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
add a comment |Â
up vote
2
down vote
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 $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
answered Sep 8 at 20:43
gimusi
71.4k73786
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
add a comment |Â
up vote
2
down vote
up vote
2
down vote
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 $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
answered Sep 8 at 20:43
gimusi
71.4k73786
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 $forall kin[0,2017]$
$$large z=r^1/2018e^icdotfractheta+2kpin$$
answered Sep 8 at 20:43
gimusi
71.4k73786
edited Sep 8 at 20:48
answered Sep 8 at 20:43
gimusi
71.4k73786
answered Sep 8 at 20:43
gimusi
71.4k73786
answered Sep 8 at 20:43
gimusi
71.4k73786
71.4k73786
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
add a comment |Â
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
Ok I got it. Sorry you have to do it again. Thanks for clearing my doubt.
– jayant98
Sep 8 at 20:57
1
1
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
@jayant98 You are welcome! Bye
– gimusi
Sep 8 at 20:58
add a comment |Â
up vote
0
down vote
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.
answered Sep 8 at 21:03
egreg
166k1180187
add a comment |Â
up vote
0
down vote
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.
answered Sep 8 at 21:03
egreg
166k1180187
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered Sep 8 at 21:03
egreg
166k1180187
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.
answered Sep 8 at 21:03
egreg
166k1180187
edited Sep 8 at 21:06
answered Sep 8 at 21:03
egreg
166k1180187
answered Sep 8 at 21:03
egreg
166k1180187
answered Sep 8 at 21:03
egreg
166k1180187
166k1180187
add a comment |Â
add a comment |Â
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Sorry, just started to learn the latex use.
– jayant98
Sep 8 at 20:42
There are 2018 complex solutions, no real solutions.
– peterh
Sep 8 at 20:58
@peterh yes you are right.
– jayant98
Sep 8 at 21:01