If $z=re^itheta$, how do I prove that $|e^iz| = e^-r sin(theta)$?Complex analysis : If $z =re^itheta$, then prove that $|e^iz| =e^-rsintheta$Given that $z= cos theta + isin theta$, prove that $Releftdfracz-1z+1right=0$Proving that $x=arccos(sqrtsintheta)$ is $sin(x+iy)=costheta+isintheta$Use the Maclaurin series to prove that $e^itheta = cos(theta) + isin(theta)$prove that $Z_1 = icot fractheta2$ given that $Z = cos theta + isin theta$Prove that $frac1+sintheta+icostheta1+sintheta-icostheta=sintheta+icostheta$Simplifying $left(frac1+sintheta + i costheta1 + sintheta -icos thetaright)^n$How to prove $1+cos2theta+cos4theta+cos6theta+cos8theta=frac(cos4theta)(sin5theta)sintheta $?Prove that $frac1+sintheta + icostheta1+sintheta-icostheta=sintheta+icostheta.$If $z=costheta + isin theta$ prove $fracz^2-1z^2+1=itantheta$
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If $z=re^itheta$, how do I prove that $|e^iz| = e^-r sin(theta)$?
Complex analysis : If $z =re^itheta$, then prove that $|e^iz| =e^-rsintheta$Given that $z= cos theta + isin theta$, prove that $Releftdfracz-1z+1right=0$Proving that $x=arccos(sqrtsintheta)$ is $sin(x+iy)=costheta+isintheta$Use the Maclaurin series to prove that $e^itheta = cos(theta) + isin(theta)$prove that $Z_1 = icot fractheta2$ given that $Z = cos theta + isin theta$Prove that $frac1+sintheta+icostheta1+sintheta-icostheta=sintheta+icostheta$Simplifying $left(frac1+sintheta + i costheta1 + sintheta -icos thetaright)^n$How to prove $1+cos2theta+cos4theta+cos6theta+cos8theta=frac(cos4theta)(sin5theta)sintheta $?Prove that $frac1+sintheta + icostheta1+sintheta-icostheta=sintheta+icostheta.$If $z=costheta + isin theta$ prove $fracz^2-1z^2+1=itantheta$
$begingroup$
If
$$
z = re^itheta
$$ how do I prove that $$left| e^iz right| = e^-r sintheta $$
solutions which I cannot understand.
I cannot understand the answer.
complex-numbers
edited Mar 21 at 7:28
Jair Taylor
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
$endgroup$
add a comment |
$begingroup$
If
$$
z = re^itheta
$$ how do I prove that $$left| e^iz right| = e^-r sintheta $$
solutions which I cannot understand.
I cannot understand the answer.
complex-numbers
edited Mar 21 at 7:28
Jair Taylor
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
$endgroup$
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31
add a comment |
$begingroup$
If
$$
z = re^itheta
$$ how do I prove that $$left| e^iz right| = e^-r sintheta $$
solutions which I cannot understand.
I cannot understand the answer.
complex-numbers
edited Mar 21 at 7:28
Jair Taylor
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
$endgroup$
If
$$
z = re^itheta
$$ how do I prove that $$left| e^iz right| = e^-r sintheta $$
solutions which I cannot understand.
I cannot understand the answer.
complex-numbers
complex-numbers
edited Mar 21 at 7:28
Jair Taylor
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
edited Mar 21 at 7:28
Jair Taylor
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
edited Mar 21 at 7:28
Jair Taylor
9,19432144
edited Mar 21 at 7:28
Jair Taylor
9,19432144
edited Mar 21 at 7:28
Jair Taylor
9,19432144
9,19432144
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
asked Mar 21 at 7:18
Aryan PandeyAryan Pandey
13
13
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31
add a comment |
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Writing $z=re^itheta= r(costheta + isintheta)$, it follows that
$$e^iz = e^ir costheta - rsintheta = e^-rsinthetae^ircostheta.$$
Recalling that $lvert e^ialpha rvert = 1 forall alpha in mathbbR$ and that $e^x > 0 forall x in mathbbR$, it follows that
$$lvert e^iz rvert = lvert e^-rsintheta rvert lvert e^ircostheta rvert = e^-rsintheta.$$
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
add a comment |
$begingroup$
Just change z in the second expresion for its value as $r (cos theta + i sin theta)$
Now, $e^ir (cos theta + i sin theta) = e^-r sin theta + ir cos theta$ The module of that is just $e^-r sin theta$ since $e^ix$ when $xin mathbbR$ is just a rotation, with no effects on module.
answered Mar 21 at 7:39
asdfasdf
2374
$endgroup$
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
$begingroup$
Writing $z=re^itheta= r(costheta + isintheta)$, it follows that
$$e^iz = e^ir costheta - rsintheta = e^-rsinthetae^ircostheta.$$
Recalling that $lvert e^ialpha rvert = 1 forall alpha in mathbbR$ and that $e^x > 0 forall x in mathbbR$, it follows that
$$lvert e^iz rvert = lvert e^-rsintheta rvert lvert e^ircostheta rvert = e^-rsintheta.$$
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
add a comment |
$begingroup$
Writing $z=re^itheta= r(costheta + isintheta)$, it follows that
$$e^iz = e^ir costheta - rsintheta = e^-rsinthetae^ircostheta.$$
Recalling that $lvert e^ialpha rvert = 1 forall alpha in mathbbR$ and that $e^x > 0 forall x in mathbbR$, it follows that
$$lvert e^iz rvert = lvert e^-rsintheta rvert lvert e^ircostheta rvert = e^-rsintheta.$$
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
$endgroup$
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
add a comment |
$begingroup$
Writing $z=re^itheta= r(costheta + isintheta)$, it follows that
$$e^iz = e^ir costheta - rsintheta = e^-rsinthetae^ircostheta.$$
Recalling that $lvert e^ialpha rvert = 1 forall alpha in mathbbR$ and that $e^x > 0 forall x in mathbbR$, it follows that
$$lvert e^iz rvert = lvert e^-rsintheta rvert lvert e^ircostheta rvert = e^-rsintheta.$$
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
$endgroup$
Writing $z=re^itheta= r(costheta + isintheta)$, it follows that
$$e^iz = e^ir costheta - rsintheta = e^-rsinthetae^ircostheta.$$
Recalling that $lvert e^ialpha rvert = 1 forall alpha in mathbbR$ and that $e^x > 0 forall x in mathbbR$, it follows that
$$lvert e^iz rvert = lvert e^-rsintheta rvert lvert e^ircostheta rvert = e^-rsintheta.$$
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
answered Mar 21 at 7:33
Gary MoonGary Moon
92127
92127
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
add a comment |
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
Wow! We answered at the same time
$endgroup$
– asdf
Mar 21 at 7:40
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
$begingroup$
It seems we did indeed! :)
$endgroup$
– Gary Moon
Mar 21 at 7:47
add a comment |
$begingroup$
Just change z in the second expresion for its value as $r (cos theta + i sin theta)$
Now, $e^ir (cos theta + i sin theta) = e^-r sin theta + ir cos theta$ The module of that is just $e^-r sin theta$ since $e^ix$ when $xin mathbbR$ is just a rotation, with no effects on module.
answered Mar 21 at 7:39
asdfasdf
2374
$endgroup$
add a comment |
$begingroup$
Just change z in the second expresion for its value as $r (cos theta + i sin theta)$
Now, $e^ir (cos theta + i sin theta) = e^-r sin theta + ir cos theta$ The module of that is just $e^-r sin theta$ since $e^ix$ when $xin mathbbR$ is just a rotation, with no effects on module.
answered Mar 21 at 7:39
asdfasdf
2374
$endgroup$
add a comment |
$begingroup$
Just change z in the second expresion for its value as $r (cos theta + i sin theta)$
Now, $e^ir (cos theta + i sin theta) = e^-r sin theta + ir cos theta$ The module of that is just $e^-r sin theta$ since $e^ix$ when $xin mathbbR$ is just a rotation, with no effects on module.
answered Mar 21 at 7:39
asdfasdf
2374
$endgroup$
Just change z in the second expresion for its value as $r (cos theta + i sin theta)$
Now, $e^ir (cos theta + i sin theta) = e^-r sin theta + ir cos theta$ The module of that is just $e^-r sin theta$ since $e^ix$ when $xin mathbbR$ is just a rotation, with no effects on module.
answered Mar 21 at 7:39
asdfasdf
2374
answered Mar 21 at 7:39
asdfasdf
2374
answered Mar 21 at 7:39
asdfasdf
2374
answered Mar 21 at 7:39
asdfasdf
2374
2374
add a comment |
add a comment |
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$begingroup$
Please see math.meta.stackexchange.com/questions/5020
$endgroup$
– Lord Shark the Unknown
Mar 21 at 7:22
$begingroup$
You can show that $left|e^wright| = e^operatornameRe(w)$ for all complex $w$. To do this, start with $w = a+bi$ and use exponential laws and results about moduli.
$endgroup$
– Minus One-Twelfth
Mar 21 at 7:31