Mixing
Calculation involving complex numbers and modules
0
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I have been racking my brain about the following for an hour now, to no avail… Can anyone help out, or at least tell me the first couple of steps in resolving this?
$z$ is a complex number and $n geq 1$ is a natural number. It is given that |z|=2. Calculate the number $α=|z^n+overline z^n|^2+|z^n-overline z^n|^2$.
Thank you!
complex-numbers
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
$endgroup$
|
show 2 more comments
0
$begingroup$
I have been racking my brain about the following for an hour now, to no avail… Can anyone help out, or at least tell me the first couple of steps in resolving this?
$z$ is a complex number and $n geq 1$ is a natural number. It is given that |z|=2. Calculate the number $α=|z^n+overline z^n|^2+|z^n-overline z^n|^2$.
Thank you!
complex-numbers
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
$endgroup$
$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
1
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53
|
show 2 more comments
0
0
0
2
$begingroup$
I have been racking my brain about the following for an hour now, to no avail… Can anyone help out, or at least tell me the first couple of steps in resolving this?
$z$ is a complex number and $n geq 1$ is a natural number. It is given that |z|=2. Calculate the number $α=|z^n+overline z^n|^2+|z^n-overline z^n|^2$.
Thank you!
complex-numbers
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
$endgroup$
I have been racking my brain about the following for an hour now, to no avail… Can anyone help out, or at least tell me the first couple of steps in resolving this?
$z$ is a complex number and $n geq 1$ is a natural number. It is given that |z|=2. Calculate the number $α=|z^n+overline z^n|^2+|z^n-overline z^n|^2$.
Thank you!
complex-numbers
complex-numbers
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
edited Jan 27 at 22:40
Eric Wofsey
190k14216349
190k14216349
asked Jan 27 at 22:34
daltadalta
1508
asked Jan 27 at 22:34
daltadalta
1508
asked Jan 27 at 22:34
daltadalta
1508
1508
$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
1
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53
|
show 2 more comments
$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
1
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53
$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
1
1
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53
|
show 2 more comments
0
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$begingroup$
Those seem reminiscent of the hyperbolic trig functions cosh and sinh. Have you tried something in that direction?
$endgroup$
– Fede Poncio
Jan 27 at 22:50
$begingroup$
Looks more like the ordinary trig functions to me, @Fede.
$endgroup$
– Gerry Myerson
Jan 27 at 23:15
$begingroup$
Yes, that’s what was on my mind. Sorry for the confusion!
$endgroup$
– Fede Poncio
Jan 28 at 0:20
$begingroup$
@FedePoncio Yes, the trig functions should be the idea, but I simply cannot work it out...
$endgroup$
– dalta
Jan 28 at 6:40
1
$begingroup$
@dalta it’s beacause $|z|=2$, so in polar coordinates then, $r=2$
$endgroup$
– Fede Poncio
Jan 28 at 16:53