Find the limit of complex number
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I have $$Z_n = e^{-i({frac{pi}{2}+{frac{1}{2n}}})}$$
Therefore, as $n to infty$,
$$operatorname{lim}Z_n = e^{-i{frac{pi}{2}}}$$
But the answer on the book is $i.$
limits complex-numbers
asked Nov 15 at 5:32
user3132457
546
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up vote
1
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I have $$Z_n = e^{-i({frac{pi}{2}+{frac{1}{2n}}})}$$
Therefore, as $n to infty$,
$$operatorname{lim}Z_n = e^{-i{frac{pi}{2}}}$$
But the answer on the book is $i.$
limits complex-numbers
asked Nov 15 at 5:32
user3132457
546
1
$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have $$Z_n = e^{-i({frac{pi}{2}+{frac{1}{2n}}})}$$
Therefore, as $n to infty$,
$$operatorname{lim}Z_n = e^{-i{frac{pi}{2}}}$$
But the answer on the book is $i.$
limits complex-numbers
asked Nov 15 at 5:32
user3132457
546
I have $$Z_n = e^{-i({frac{pi}{2}+{frac{1}{2n}}})}$$
Therefore, as $n to infty$,
$$operatorname{lim}Z_n = e^{-i{frac{pi}{2}}}$$
But the answer on the book is $i.$
limits complex-numbers
limits complex-numbers
asked Nov 15 at 5:32
user3132457
546
asked Nov 15 at 5:32
user3132457
546
asked Nov 15 at 5:32
user3132457
546
asked Nov 15 at 5:32
user3132457
546
asked Nov 15 at 5:32
user3132457
546
546
1
$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39
add a comment |
1
$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39
1
1
$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39
$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39
add a comment |
1 Answer
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Note that $$e^{-ipi /2} = cos (-pi/2) -i sin(pi/2)=-i$$
Thus the book has a typo.
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $$e^{-ipi /2} = cos (-pi/2) -i sin(pi/2)=-i$$
Thus the book has a typo.
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
add a comment |
up vote
1
down vote
accepted
Note that $$e^{-ipi /2} = cos (-pi/2) -i sin(pi/2)=-i$$
Thus the book has a typo.
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $$e^{-ipi /2} = cos (-pi/2) -i sin(pi/2)=-i$$
Thus the book has a typo.
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
Note that $$e^{-ipi /2} = cos (-pi/2) -i sin(pi/2)=-i$$
Thus the book has a typo.
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
answered Nov 15 at 5:43
Mohammad Riazi-Kermani
39.8k41957
39.8k41957
add a comment |
add a comment |
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$dfrac1nto0$ so the answer is $-i$.
– Nosrati
Nov 15 at 5:39