(2i\1-i)²=

计算:(根号2/(1+i))^100+ (根号2/(1-i))^1000分_百度知道
计算:(根号2/(1+i))^100+ (根号2/(1-i))^1000分
(根号2/(1+i))^100+ (根号2/(1-i))^100
提问者采纳
(1-i)^100=2^50/)^25=-2^50(1-i)^100=[(1-i)²]^50=(1+i&#178(1+i)^100=[(1+i)²(1+i))^100+ (根号2/(1+i)^100+ 2^50/]^50=(1+i²]^25=-4^25=-(2²(1-i))^100=2^50/+2i)^50=[(2i)²-2i)^50=[(-2i)²(-2^50) + 2^50/)^25=-2^50(根号2/]^25=-4^25=-(2&#178
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出门在外也不愁csc480°= 1-2i/3+i= 1\csc480°=2\1-2i/3+i=3\Given a1=1,an=2an-1(见图片),then the sum of the first 10 terms of the sequence is 4\If the center of circle x²+y²+ax+by-5=0 is(1,1/2),then the radius is 5\The standard equation of a parab_百度作业帮
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csc480°= 1-2i/3+i= 1\csc480°=2\1-2i/3+i=3\Given a1=1,an=2an-1(见图片),then the sum of the first 10 terms of the sequence is 4\If the center of circle x²+y²+ax+by-5=0 is(1,1/2),then the radius is 5\The standard equation of a parab
csc480°= 1-2i/3+i= 1\csc480°=2\1-2i/3+i=3\Given a1=1,an=2an-1(见图片),then the sum of the first 10 terms of the sequence is 4\If the center of circle x²+y²+ax+by-5=0 is(1,1/2),then the radius is 5\The standard equation of a parabola with focus(1,2),and directrix y=6 is 6\In△ABC,AC=1,AB=根号3,角B=30°,then the area of △ABCis?7\Given(x-1)²-16(y+2)²=16,find the center,thefocus,the vertex,the transverse axis lenth and the conjugate axis length,the eccentricity,the asymptotes.file:///C:/Documents%20and%20Settings/Administrator/%E6%A1%8C%E9%9D%A2/%E6%9C%AA%E5%91%BD%E5%90%8D2.JPG
csc480°= csc120°=1/sin120°=1/(√3/2)=2/√3=2√3/3 1-2i/3+i= (1-2i)(3-i)/(3+i)(3-i)=(3-7i-2)/10=(1-7i)/10设i为虚数单位,则1+i+i2+i3+···+i^2014=?_百度知道
设i为虚数单位,则1+i+i2+i3+···+i^2014=?
提问者采纳
+i^2014=1·(1-i^2015)&#47.+i^2014=1·(1-i^2015)/(1-i)=(1+i)/+;&#47.可以看做是以1为首项.;(1-i)i^2015=i·(i²)^1007=i·(-1)^1007=-i1+i+i²+..;(1-i)=(1+i)&#178,i为公比的等比数列;(1-i)=1·[1-(-i)]&#47。1+i+i&#178
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答..;+···+i^2014=1+i+i&#178:i^0=1i^1=ii²=1+i-1-i=0∴1+i+i&#178,请点击“采纳为满意答案”;=-ii^4=1i^5=ii^6=-1i^7=-i。;+···+i^2014共2015项2015/=1+i-1=i如果您认可我的回答.。.;+i&#179.每四项一循环1+i+i&#178。3∴1+i+i²+i³4=503;+i³=-1i&#179,祝学习进步
1+i+i2+i3+···+i^+3+...+2014)i=1+(2+i/2=1+2013^2i
认真看题目是 1+i+i²+i³+···+i^2014=?
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出门在外也不愁在复数集内解方程:(x^2-2ix-5)(x^3+i)(x^3-2x^2+x-2)=0_百度知道
在复数集内解方程:(x^2-2ix-5)(x^3+i)(x^3-2x^2+x-2)=0
我有更好的答案
-4=0或x+i=0或(x+i/2或x=0或x=±i;-4)(x+i)((x+i&#47(x^2-2ix-5)(x^3+i)(x^3-2x^2+x-2)=0((x-i)&#178,如果满意记得采纳如果有其他问题请采纳本题后另发点击向我求助;∴x=i±2或x=-i或x=(-i±√5i)/+1)=0((x-i)²+5/4=0或x-2=0或x&#178,谢谢; 如果本题有什么不明白可以追问;+1=0,请谅解;+5/-4)(x+i)(x²2)²2)&#178,答题不易;4)(x-2)(x²+1)=0∴(x-i)²+ix+1)(x-2)(x&#178
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