Simple way to solve the product of a imaginary number's exponent:

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I came up with is to take the imaginary number's exponent, divide it by 2, and if it's positive then the answer is 1. If the result is odd then the answer is -1.



Example:



$i^8$



$8/2$



$4$



Since 4 is even the answer to $i^8$ is 1.






algebra-precalculus







share|cite|improve this question























  • How would that work for $,i^-3,$, for example?
    – dxiv
    Sep 7 at 7:21











  • I think my formula doesn't really work :(
    – Samurai
    Sep 7 at 7:33














up vote
0
down vote

favorite












I came up with is to take the imaginary number's exponent, divide it by 2, and if it's positive then the answer is 1. If the result is odd then the answer is -1.



Example:



$i^8$



$8/2$



$4$



Since 4 is even the answer to $i^8$ is 1.






algebra-precalculus







share|cite|improve this question























  • How would that work for $,i^-3,$, for example?
    – dxiv
    Sep 7 at 7:21











  • I think my formula doesn't really work :(
    – Samurai
    Sep 7 at 7:33












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I came up with is to take the imaginary number's exponent, divide it by 2, and if it's positive then the answer is 1. If the result is odd then the answer is -1.



Example:



$i^8$



$8/2$



$4$



Since 4 is even the answer to $i^8$ is 1.






algebra-precalculus







share|cite|improve this question















I came up with is to take the imaginary number's exponent, divide it by 2, and if it's positive then the answer is 1. If the result is odd then the answer is -1.



Example:



$i^8$



$8/2$



$4$



Since 4 is even the answer to $i^8$ is 1.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 7 at 7:36









N. F. Taussig

39.7k93153




39.7k93153










asked Sep 7 at 7:18









Samurai

32




32











  • How would that work for $,i^-3,$, for example?
    – dxiv
    Sep 7 at 7:21











  • I think my formula doesn't really work :(
    – Samurai
    Sep 7 at 7:33
















  • How would that work for $,i^-3,$, for example?
    – dxiv
    Sep 7 at 7:21











  • I think my formula doesn't really work :(
    – Samurai
    Sep 7 at 7:33















How would that work for $,i^-3,$, for example?
– dxiv
Sep 7 at 7:21





How would that work for $,i^-3,$, for example?
– dxiv
Sep 7 at 7:21













I think my formula doesn't really work :(
– Samurai
Sep 7 at 7:33




I think my formula doesn't really work :(
– Samurai
Sep 7 at 7:33










1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










Let us stick to positive exponents for now:



Since $i^0 = 1$, work out $i^1, i^2, i^3$, and $i^4$, and then you will notice a pattern.



The answer to $i^4$ can be justified, since multiplying by $i$ is the same as rotating by $90º$ (in the counter-clockwise direction).






share|cite|improve this answer




















  • I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
    – Samurai
    Sep 7 at 17:41










  • @Samurai Thanks for the accept!
    – Toby Mak
    Sep 8 at 0:00










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908348%2fsimple-way-to-solve-the-product-of-a-imaginary-numbers-exponent%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Let us stick to positive exponents for now:



Since $i^0 = 1$, work out $i^1, i^2, i^3$, and $i^4$, and then you will notice a pattern.



The answer to $i^4$ can be justified, since multiplying by $i$ is the same as rotating by $90º$ (in the counter-clockwise direction).






share|cite|improve this answer




















  • I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
    – Samurai
    Sep 7 at 17:41










  • @Samurai Thanks for the accept!
    – Toby Mak
    Sep 8 at 0:00














up vote
0
down vote



accepted










Let us stick to positive exponents for now:



Since $i^0 = 1$, work out $i^1, i^2, i^3$, and $i^4$, and then you will notice a pattern.



The answer to $i^4$ can be justified, since multiplying by $i$ is the same as rotating by $90º$ (in the counter-clockwise direction).






share|cite|improve this answer




















  • I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
    – Samurai
    Sep 7 at 17:41










  • @Samurai Thanks for the accept!
    – Toby Mak
    Sep 8 at 0:00












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Let us stick to positive exponents for now:



Since $i^0 = 1$, work out $i^1, i^2, i^3$, and $i^4$, and then you will notice a pattern.



The answer to $i^4$ can be justified, since multiplying by $i$ is the same as rotating by $90º$ (in the counter-clockwise direction).






share|cite|improve this answer












Let us stick to positive exponents for now:



Since $i^0 = 1$, work out $i^1, i^2, i^3$, and $i^4$, and then you will notice a pattern.



The answer to $i^4$ can be justified, since multiplying by $i$ is the same as rotating by $90º$ (in the counter-clockwise direction).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 7 at 7:35









Toby Mak

2,8751925




2,8751925











  • I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
    – Samurai
    Sep 7 at 17:41










  • @Samurai Thanks for the accept!
    – Toby Mak
    Sep 8 at 0:00
















  • I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
    – Samurai
    Sep 7 at 17:41










  • @Samurai Thanks for the accept!
    – Toby Mak
    Sep 8 at 0:00















I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
– Samurai
Sep 7 at 17:41




I see where my simple formula falls short. Every other result is a decimal and a decimal really can't be even or odd. On top of that sometimes the answer is just i.
– Samurai
Sep 7 at 17:41












@Samurai Thanks for the accept!
– Toby Mak
Sep 8 at 0:00




@Samurai Thanks for the accept!
– Toby Mak
Sep 8 at 0:00

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2908348%2fsimple-way-to-solve-the-product-of-a-imaginary-numbers-exponent%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier curve. Additionally it marks one side of the curve as inside t
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容