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Question Number 123607 by mohammad17 last updated on 26/Nov/20

if a=((1+i)/( (√2)))  find a^(1943)     ?

ifa=1+i2finda1943?

Answered by Dwaipayan Shikari last updated on 26/Nov/20

a =((1+i)/( (√2))) =(√i)  a^(1943) =(i)^((1943)/2) =i^(970) .i.(√i) =−i(√i)  Another way  a=((1+i)/( (√2)))=e^((iπ)/4)     ⇒a^(1943) =e^((1943πi)/4) =e^(435πi) e^(((3π)/4)i)  =−1.(−(1/( (√2)))+(i/( (√2))))  =(1/( (√2)))−(i/( (√2)))

a=1+i2=ia1943=(i)19432=i970.i.i=iiAnotherwaya=1+i2=eiπ4a1943=e1943πi4=e435πie3π4i=1.(12+i2)=12i2

Answered by malwan last updated on 26/Nov/20

a=[1 , (π/4)]  ⇒a^(1943)  = [1^(1943)  , 1943×(π/4)]= [1 , π] [1 , ((3π)/4)]  = −1(cos((3π)/4) + i sin ((3π)/4)) = ((1−i)/( (√2)))

a=[1,π4]a1943=[11943,1943×π4]=[1,π][1,3π4]=1(cos3π4+isin3π4)=1i2

Answered by TANMAY PANACEA last updated on 26/Nov/20

a=(1/( (√2)))+i(1/( (√2)))=cos(π/4)+isin(π/4)=e^(i×(π/4))   1943=4×485+3  a=e^(i×(π/4))   a^(1943) =(e^(i×(π/4)) )^(4×485+3) =e^(i×485π+i×((3π)/4))     e^(i×485π) ×e^(i×((3π)/4))   =(cos485π+isin485π)(cos135^o +isin135^o )  ={cos(242×2π+π)+i×0}(−(1/( (√2)))+i×(1/( (√2))))  =(cosπ+0)(((−1)/( (√2)))+i×(1/( (√2))))  =(−1)(((−1)/( (√2)))+i×(1/( (√2))))=(1/( (√2)))−(i/( (√2)))

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)=12i2

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