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Question Number 123607 by mohammad17 last updated on 26/Nov/20
if a=((1+i)/( (√2))) find a^(1943) ?
$${if}\:{a}=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:\:{find}\:{a}^{\mathrm{1943}} \:\:\:\:? \\ $$
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)))
$${a}\:=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{{i}} \\ $$$${a}^{\mathrm{1943}} =\left({i}\right)^{\frac{\mathrm{1943}}{\mathrm{2}}} ={i}^{\mathrm{970}} .{i}.\sqrt{{i}}\:=−{i}\sqrt{{i}} \\ $$$${Another}\:{way} \\ $$$${a}=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}={e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\Rightarrow{a}^{\mathrm{1943}} ={e}^{\frac{\mathrm{1943}\pi{i}}{\mathrm{4}}} ={e}^{\mathrm{435}\pi{i}} {e}^{\frac{\mathrm{3}\pi}{\mathrm{4}}{i}} \:=−\mathrm{1}.\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{{i}}{\:\sqrt{\mathrm{2}}} \\ $$
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)))
$${a}=\left[\mathrm{1}\:,\:\frac{\pi}{\mathrm{4}}\right] \\ $$$$\Rightarrow{a}^{\mathrm{1943}} \:=\:\left[\mathrm{1}^{\mathrm{1943}} \:,\:\mathrm{1943}×\frac{\pi}{\mathrm{4}}\right]=\:\left[\mathrm{1}\:,\:\pi\right]\:\left[\mathrm{1}\:,\:\frac{\mathrm{3}\pi}{\mathrm{4}}\right] \\ $$$$=\:−\mathrm{1}\left({cos}\frac{\mathrm{3}\pi}{\mathrm{4}}\:+\:{i}\:{sin}\:\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\:=\:\frac{\mathrm{1}−{i}}{\:\sqrt{\mathrm{2}}} \\ $$
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)))
$${a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}={cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}={e}^{{i}×\frac{\pi}{\mathrm{4}}} \\ $$$$\mathrm{1943}=\mathrm{4}×\mathrm{485}+\mathrm{3} \\ $$$${a}={e}^{{i}×\frac{\pi}{\mathrm{4}}} \\ $$$${a}^{\mathrm{1943}} =\left({e}^{{i}×\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{4}×\mathrm{485}+\mathrm{3}} ={e}^{{i}×\mathrm{485}\pi+{i}×\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$ \\ $$$${e}^{{i}×\mathrm{485}\pi} ×{e}^{{i}×\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$=\left({cos}\mathrm{485}\pi+{isin}\mathrm{485}\pi\right)\left({cos}\mathrm{135}^{{o}} +{isin}\mathrm{135}^{{o}} \right) \\ $$$$=\left\{{cos}\left(\mathrm{242}×\mathrm{2}\pi+\pi\right)+{i}×\mathrm{0}\right\}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+{i}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\left({cos}\pi+\mathrm{0}\right)\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}+{i}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\left(−\mathrm{1}\right)\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}+{i}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{{i}}{\:\sqrt{\mathrm{2}}} \\ $$

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