solving $cisX=cisY$ equations
Part of a solution I came across with solving the following equation: $cis(nfrac{pi}{3})=1$. I converted it to be $cis(nfrac{pi}{3})=cis0$.
After trying to insert some values, I understood that the solution is $nfrac{pi}{3}=2pi k$ when $kinmathbb{Z}$ meaning $n=6k$.
Is it true to say that the solution of $cisX=cisY$ is $X=Y+2pi k$ when $kinmathbb{Z}$?
complex-numbers
edited Nov 23 at 14:21
D...
213113
asked Nov 23 at 13:55
vesii
635
add a comment |
Part of a solution I came across with solving the following equation: $cis(nfrac{pi}{3})=1$. I converted it to be $cis(nfrac{pi}{3})=cis0$.
After trying to insert some values, I understood that the solution is $nfrac{pi}{3}=2pi k$ when $kinmathbb{Z}$ meaning $n=6k$.
Is it true to say that the solution of $cisX=cisY$ is $X=Y+2pi k$ when $kinmathbb{Z}$?
complex-numbers
edited Nov 23 at 14:21
D...
213113
asked Nov 23 at 13:55
vesii
635
add a comment |
Part of a solution I came across with solving the following equation: $cis(nfrac{pi}{3})=1$. I converted it to be $cis(nfrac{pi}{3})=cis0$.
After trying to insert some values, I understood that the solution is $nfrac{pi}{3}=2pi k$ when $kinmathbb{Z}$ meaning $n=6k$.
Is it true to say that the solution of $cisX=cisY$ is $X=Y+2pi k$ when $kinmathbb{Z}$?
complex-numbers
edited Nov 23 at 14:21
D...
213113
asked Nov 23 at 13:55
vesii
635
Part of a solution I came across with solving the following equation: $cis(nfrac{pi}{3})=1$. I converted it to be $cis(nfrac{pi}{3})=cis0$.
After trying to insert some values, I understood that the solution is $nfrac{pi}{3}=2pi k$ when $kinmathbb{Z}$ meaning $n=6k$.
Is it true to say that the solution of $cisX=cisY$ is $X=Y+2pi k$ when $kinmathbb{Z}$?
complex-numbers
complex-numbers
edited Nov 23 at 14:21
D...
213113
asked Nov 23 at 13:55
vesii
635
edited Nov 23 at 14:21
D...
213113
asked Nov 23 at 13:55
vesii
635
edited Nov 23 at 14:21
D...
213113
edited Nov 23 at 14:21
D...
213113
edited Nov 23 at 14:21
D...
213113
213113
asked Nov 23 at 13:55
vesii
635
asked Nov 23 at 13:55
vesii
635
asked Nov 23 at 13:55
vesii
635
635
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality
$cos(X) + i sin(X) = cos(Y) + i sin(Y)$
holds if and only if
$cos(X) = cos(Y)$ and $sin(X) = sin(Y)$.
Can you solve from here?
answered Nov 23 at 14:12
D...
213113
add a comment |
Your answer is completely correct but the sine function is zero twice in a period.
$$sin(X)=0$$
$$X=0+kpi=kpi k∈Z$$
so
$$X=3k text{and not} 6k$$
edited Nov 23 at 14:41
MRobinson
1,765319
answered Nov 23 at 14:20
Etotheipi
344
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality
$cos(X) + i sin(X) = cos(Y) + i sin(Y)$
holds if and only if
$cos(X) = cos(Y)$ and $sin(X) = sin(Y)$.
Can you solve from here?
answered Nov 23 at 14:12
D...
213113
add a comment |
Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality
$cos(X) + i sin(X) = cos(Y) + i sin(Y)$
holds if and only if
$cos(X) = cos(Y)$ and $sin(X) = sin(Y)$.
Can you solve from here?
answered Nov 23 at 14:12
D...
213113
add a comment |
Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality
$cos(X) + i sin(X) = cos(Y) + i sin(Y)$
holds if and only if
$cos(X) = cos(Y)$ and $sin(X) = sin(Y)$.
Can you solve from here?
answered Nov 23 at 14:12
D...
213113
Yes, it is. Two complex numbers are equal if and only if their real and imaginary parts coincide. Therefore, the equality
$cos(X) + i sin(X) = cos(Y) + i sin(Y)$
holds if and only if
$cos(X) = cos(Y)$ and $sin(X) = sin(Y)$.
Can you solve from here?
answered Nov 23 at 14:12
D...
213113
answered Nov 23 at 14:12
D...
213113
answered Nov 23 at 14:12
D...
213113
answered Nov 23 at 14:12
D...
213113
213113
add a comment |
add a comment |
Your answer is completely correct but the sine function is zero twice in a period.
$$sin(X)=0$$
$$X=0+kpi=kpi k∈Z$$
so
$$X=3k text{and not} 6k$$
edited Nov 23 at 14:41
MRobinson
1,765319
answered Nov 23 at 14:20
Etotheipi
344
add a comment |
Your answer is completely correct but the sine function is zero twice in a period.
$$sin(X)=0$$
$$X=0+kpi=kpi k∈Z$$
so
$$X=3k text{and not} 6k$$
edited Nov 23 at 14:41
MRobinson
1,765319
answered Nov 23 at 14:20
Etotheipi
344
add a comment |
Your answer is completely correct but the sine function is zero twice in a period.
$$sin(X)=0$$
$$X=0+kpi=kpi k∈Z$$
so
$$X=3k text{and not} 6k$$
edited Nov 23 at 14:41
MRobinson
1,765319
answered Nov 23 at 14:20
Etotheipi
344
Your answer is completely correct but the sine function is zero twice in a period.
$$sin(X)=0$$
$$X=0+kpi=kpi k∈Z$$
so
$$X=3k text{and not} 6k$$
edited Nov 23 at 14:41
MRobinson
1,765319
answered Nov 23 at 14:20
Etotheipi
344
edited Nov 23 at 14:41
MRobinson
1,765319
edited Nov 23 at 14:41
MRobinson
1,765319
edited Nov 23 at 14:41
MRobinson
1,765319
1,765319
answered Nov 23 at 14:20
Etotheipi
344
answered Nov 23 at 14:20
Etotheipi
344
answered Nov 23 at 14:20
Etotheipi
344
344
add a comment |
add a comment |
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