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Question Number 123607 by mohammad17 last updated on 26/Nov/20
ifa=1+i2finda1943?
Answered by Dwaipayan Shikari last updated on 26/Nov/20
a=1+i2=ia1943=(i)19432=i970.i.i=−iiAnotherwaya=1+i2=eiπ4⇒a1943=e1943πi4=e435πie3π4i=−1.(−12+i2)=12−i2
Answered by malwan last updated on 26/Nov/20
a=[1,π4]⇒a1943=[11943,1943×π4]=[1,π][1,3π4]=−1(cos3π4+isin3π4)=1−i2
Answered by TANMAY PANACEA last updated on 26/Nov/20
a=12+i12=cosπ4+isinπ4=ei×π41943=4×485+3a=ei×π4a1943=(ei×π4)4×485+3=ei×485π+i×3π4ei×485π×ei×3π4=(cos485π+isin485π)(cos135o+isin135o)={cos(242×2π+π)+i×0}(−12+i×12)=(cosπ+0)(−12+i×12)=(−1)(−12+i×12)=12−i2
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