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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [\$\\sin(2\\pi i) = ?\$ (imaginary '\$i\$' inside the paranthesis) \[closed\]](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 8 years, 3 months ago Modified [8 years, 3 months ago](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis?lastactivity "2017-11-02 07:46:10Z") Viewed 4k times \-3 \$\\begingroup\$ **Closed.** This question is [off-topic](https://math.stackexchange.com/help/closed-questions). It is not currently accepting answers. *** **This question is missing context or other details**: Please [improve the question](https://math.meta.stackexchange.com/questions/9959) by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Closed 8 years ago. [Improve this question](https://math.stackexchange.com/posts/2500934/edit) \$\\sin(2\\pi i)\$ equals what ? I tried Euler's formula and got \$-267.74i\$ , i.e. \$\$\\frac{e^{2\\pi i} - e^{-2\\pi i}}{2i}\$\$ but I don't think it would be right, because \$\|\\sin x\| \\leq 1\$ always, right ? So, how to find this value. (question by IISc in a competitive exam) - [complex-analysis](https://math.stackexchange.com/questions/tagged/complex-analysis "show questions tagged 'complex-analysis'") [Share](https://math.stackexchange.com/q/2500934 "Short permalink to this question") Cite Follow [edited Nov 2, 2017 at 7:46](https://math.stackexchange.com/posts/2500934/revisions "show all edits to this post") [](https://math.stackexchange.com/users/460274/teddy38) [Teddy38](https://math.stackexchange.com/users/460274/teddy38) 3,36322 gold badges1414 silver badges3333 bronze badges asked Nov 2, 2017 at 6:43 [](https://math.stackexchange.com/users/498402/jayanth) [Jayanth](https://math.stackexchange.com/users/498402/jayanth) 511 silver badge33 bronze badges \$\\endgroup\$ [Add a comment](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \| ## 3 Answers 3 Sorted by: [Reset to default](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis?answertab=scoredesc#tab-top) 2 \$\\begingroup\$ \$\|\\sin(x) \\le 1\|\$ is true on the real line. On the imaginary axis \$\\sin\$ behaves llike \$\\sinh\$ on the real line. [Share](https://math.stackexchange.com/a/2500935 "Short permalink to this answer") Cite Follow answered Nov 2, 2017 at 6:46 [](https://math.stackexchange.com/users/128832/thomas) [Thomas](https://math.stackexchange.com/users/128832/thomas) 23\.3k22 gold badges3131 silver badges4444 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ Thank you Thomas (really quick reply). So, would 267.75 be right ? (sin(2\*pi\*i) = 267.75) ? \$\\endgroup\$ Jayanth – [Jayanth](https://math.stackexchange.com/users/498402/jayanth "5 reputation") 2017-11-02 06:50:23 +00:00 Commented Nov 2, 2017 at 6:50 [Add a comment](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| 1 \$\\begingroup\$ \$\$\\sin(ix)=i\\sinh(x)\$\$ so \$\$\\sin(2\\pi i)=i\\sinh(2\\pi)\$\$ which is approximately \$267\\cdot7i\$. The inequality \$\|\\sin x\|\\le1\$ is **not** valid when \$x\$ is non-real. [Share](https://math.stackexchange.com/a/2500938 "Short permalink to this answer") Cite Follow [edited Nov 2, 2017 at 7:00](https://math.stackexchange.com/posts/2500938/revisions "show all edits to this post") answered Nov 2, 2017 at 6:47 [](https://math.stackexchange.com/users/436618/angina-seng) [Angina Seng](https://math.stackexchange.com/users/436618/angina-seng) 163k2828 gold badges117117 silver badges221221 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ thank you brother, :) Helped. \$\\endgroup\$ Jayanth – [Jayanth](https://math.stackexchange.com/users/498402/jayanth "5 reputation") 2017-11-02 07:01:39 +00:00 Commented Nov 2, 2017 at 7:01 [Add a comment](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| \-1 \$\\begingroup\$ We need to use \$\$2i\\sin x=e^{ix}-e^{-ix}\$\$ Now \$\\sin^2z+\\cos^2z=1\$ holds true for all complex/real \$z\$ \$\$\\cos^2z=1-\\sin^2z\$\$ Now only for real \$\\cos z,\\cos^2z\\ge0\$ consequently, we need \$\$1-\\sin^2z\\ge0\\iff-1\\le\\sin z\\le1\$\$ [Share](https://math.stackexchange.com/a/2500936 "Short permalink to this answer") Cite Follow answered Nov 2, 2017 at 6:47 [](https://math.stackexchange.com/users/33337/lab-bhattacharjee) [lab bhattacharjee](https://math.stackexchange.com/users/33337/lab-bhattacharjee) 280k2121 gold badges213213 silver badges338338 bronze badges \$\\endgroup\$ 2 - \$\\begingroup\$ Wish I know the mistake here \$\\endgroup\$ lab bhattacharjee – [lab bhattacharjee](https://math.stackexchange.com/users/33337/lab-bhattacharjee "280,452 reputation") 2017-11-02 08:54:46 +00:00 Commented Nov 2, 2017 at 8:54 - 1 \$\\begingroup\$ I didn't downvote, but 1. You seem to ignore the fact the OP knows how to get sine from Euler's Formula. 2. The phrase "we need" gives the impression that the OP's misconception is correct, especially since you end on an inequality implying sine is real. It's true that you say in the middle "only for real cos \$z \$", but this could be lost in the wording of your answer. 3. This is more minor, but you dont support your claims about real cosine, or the pythagorean identity, or the implicit claim that cosine can be nonreal. \$\\endgroup\$ Mark S. – [Mark S.](https://math.stackexchange.com/users/26369/mark-s "26,486 reputation") 2017-11-02 10:02:07 +00:00 Commented Nov 2, 2017 at 10:02 [Add a comment](https://math.stackexchange.com/questions/2500934/sin2-pi-i-imaginary-i-inside-the-paranthesis "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| Start asking to get answers Find the answer to your question by asking. 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\-3 \$\\begingroup\$ \$\\sin(2\\pi i)\$ equals what ? I tried Euler's formula and got \$-267.74i\$ , i.e. \$\$\\frac{e^{2\\pi i} - e^{-2\\pi i}}{2i}\$\$ but I don't think it would be right, because \$\|\\sin x\| \\leq 1\$ always, right ? So, how to find this value. (question by IISc in a competitive exam) [](https://math.stackexchange.com/users/460274/teddy38) [Teddy38](https://math.stackexchange.com/users/460274/teddy38) 3,3632 gold badges14 silver badges33 bronze badges asked Nov 2, 2017 at 6:43 [](https://math.stackexchange.com/users/498402/jayanth) \$\\endgroup\$ 2 \$\\begingroup\$ \$\|\\sin(x) \\le 1\|\$ is true on the real line. On the imaginary axis \$\\sin\$ behaves llike \$\\sinh\$ on the real line. answered Nov 2, 2017 at 6:46 [](https://math.stackexchange.com/users/128832/thomas) \$\\endgroup\$ 1 1 \$\\begingroup\$ \$\$\\sin(ix)=i\\sinh(x)\$\$ so \$\$\\sin(2\\pi i)=i\\sinh(2\\pi)\$\$ which is approximately \$267\\cdot7i\$. The inequality \$\|\\sin x\|\\le1\$ is **not** valid when \$x\$ is non-real. answered Nov 2, 2017 at 6:47 [](https://math.stackexchange.com/users/436618/angina-seng) \$\\endgroup\$ 1 \-1 \$\\begingroup\$ We need to use \$\$2i\\sin x=e^{ix}-e^{-ix}\$\$ Now \$\\sin^2z+\\cos^2z=1\$ holds true for all complex/real \$z\$ \$\$\\cos^2z=1-\\sin^2z\$\$ Now only for real \$\\cos z,\\cos^2z\\ge0\$ consequently, we need \$\$1-\\sin^2z\\ge0\\iff-1\\le\\sin z\\le1\$\$ answered Nov 2, 2017 at 6:47 [](https://math.stackexchange.com/users/33337/lab-bhattacharjee) \$\\endgroup\$ 2 Start asking to get answers Find the answer to your question by asking. [Ask question](https://math.stackexchange.com/questions/ask) Explore related questions See similar questions with these tags.