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Question Number 221697 by fantastic last updated on 09/Jun/25
Is (√i) an imaginary number (i=(√(−1))) answer with logic
$${Is}\:\sqrt{{i}}\:{an}\:{imaginary}\:{number}\:\left({i}=\sqrt{−\mathrm{1}}\right)\:{answer}\:{with}\:{logic} \\ $$
Answered by Ghisom last updated on 09/Jun/25
i=e^(i(π/2)) z=re^(iθ) ⇒ (√z)=(√r)e^(i(θ/2)) ⇒ (√i)=e^(i(π/4)) =((√2)/2)+((√2)/2)i
$$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\sqrt{\mathrm{i}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$
Answered by wewji12 last updated on 09/Jun/25
From....Euler′s identity e^(it) =cos(t)+isin(t) e^z =Σ_(h=0) ^∞ (z^h /(h!)) e^(iz) =Σ_(h=0) ^∞ (((iz)^h )/(h!))=1+iz−(z^2 /(2!))−((iz^3 )/(3!))+(z^4 /(4!))+((iz^5 )/(5!))..... Real{e^(iz) }=1−(z^2 /(2!))+(z^4 /(4!))...= Tayler Series of cos(z) imag{e^(iz) }=z−(z^3 /(3!))+(z^5 /(5!))...=Tayler Series of sin(z) So we can get e^(it) =cos(t)+i∙sin(t) and Let′s substitute t=πt′ e^(iπt′) =cos(πt′)+isin(πt′) e^((1/4)iπ) =cos((1/4)π)+i∙sin((1/4)π) ∵ e^((1/2)iπ) =i →e^((1/4)iπ) =(√i) So... (√i)=(1/( (√2)))(1+i)
$$\mathrm{From}….\mathrm{Euler}'\mathrm{s}\:\mathrm{identity} \\ $$$${e}^{\boldsymbol{{i}}{t}} =\mathrm{cos}\left({t}\right)+\boldsymbol{{i}}\mathrm{sin}\left({t}\right) \\ $$$${e}^{{z}} =\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{{z}^{{h}} }{{h}!}\: \\ $$$${e}^{\boldsymbol{{i}}{z}} =\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(\boldsymbol{{i}}{z}\right)^{{h}} }{{h}!}=\mathrm{1}+\boldsymbol{{i}}{z}−\frac{{z}^{\mathrm{2}} }{\mathrm{2}!}−\frac{\boldsymbol{{i}}{z}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{z}^{\mathrm{4}} }{\mathrm{4}!}+\frac{\boldsymbol{{i}}{z}^{\mathrm{5}} }{\mathrm{5}!}….. \\ $$$$\mathrm{Real}\left\{{e}^{\boldsymbol{{i}}{z}} \right\}=\mathrm{1}−\frac{{z}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{z}^{\mathrm{4}} }{\mathrm{4}!}…=\:\mathrm{Tayler}\:\mathrm{Series}\:\mathrm{of}\:\mathrm{cos}\left({z}\right)\: \\ $$$$\mathrm{imag}\left\{{e}^{\boldsymbol{{i}}{z}} \right\}={z}−\frac{{z}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{z}^{\mathrm{5}} }{\mathrm{5}!}…=\mathrm{Tayler}\:\mathrm{Series}\:\mathrm{of}\:\mathrm{sin}\left({z}\right) \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:{e}^{\boldsymbol{{i}}{t}} =\mathrm{cos}\left({t}\right)+\boldsymbol{{i}}\centerdot\mathrm{sin}\left({t}\right) \\ $$$$\mathrm{and}\:\mathrm{Let}'\mathrm{s}\:\mathrm{substitute}\:{t}=\pi{t}' \\ $$$${e}^{\boldsymbol{{i}}\pi{t}'} =\mathrm{cos}\left(\pi{t}'\right)+\boldsymbol{{i}}\mathrm{sin}\left(\pi{t}'\right) \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{i}}\pi} =\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\right)+\boldsymbol{{i}}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\right)\:\because\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{i}}\pi} =\boldsymbol{{i}}\:\rightarrow{e}^{\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{i}}\pi} =\sqrt{\boldsymbol{{i}}} \\ $$$$\mathrm{So}…\:\sqrt{\boldsymbol{{i}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+\boldsymbol{{i}}\right) \\ $$
Commented by fantastic last updated on 09/Jun/25
just write (1/( (√2)))((√(2i)))=(1/( (√2)))(√(1+2i+i^2 ))=(1/( (√2)))(1+i)
$${just}\:{write}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{\mathrm{2}{i}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\mathrm{1}+\mathrm{2}{i}+{i}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$
Commented by wewji12 last updated on 09/Jun/25
No..... Your answer is not wrong. but it does not guarantee matematical rigor Of course not all of my answers are correct because complex functions are multivalued function.. so they have so many values(almost infinity Solutions) so to be more precise we have to use techniques from complex function theories liked Monoromy and Brach Cut...
$$\mathrm{No}….. \\ $$$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{wrong}.\:\mathrm{but}\:\mathrm{it}\:\mathrm{does}\:\mathrm{not}\:\mathrm{guarantee} \\ $$$$\mathrm{matematical}\:\mathrm{rigor} \\ $$$$\mathrm{Of}\:\mathrm{course}\:\mathrm{not}\:\mathrm{all}\:\mathrm{of}\:\mathrm{my}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{correct}\:\mathrm{because} \\ $$$$\mathrm{complex}\:\mathrm{functions}\:\mathrm{are}\:\mathrm{multivalued}\:\mathrm{function}.. \\ $$$$\mathrm{so}\:\mathrm{they}\:\mathrm{have}\:\mathrm{so}\:\mathrm{many}\:\mathrm{values}\left(\mathrm{almost}\:\mathrm{infinity}\:\mathrm{Solutions}\right) \\ $$$$\mathrm{so}\:\mathrm{to}\:\mathrm{be}\:\mathrm{more}\:\mathrm{precise}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to} \\ $$$$\mathrm{use}\:\mathrm{techniques}\:\mathrm{from}\:\mathrm{complex}\:\mathrm{function}\:\mathrm{theories}\:\mathrm{liked} \\ $$$$\mathrm{Monoromy}\:\mathrm{and}\:\mathrm{Brach}\:\mathrm{Cut}… \\ $$
Commented by fantastic last updated on 09/Jun/25
right sir
$${right}\:{sir} \\ $$

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