Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 82041 by M±th+et£s last updated on 17/Feb/20

$$showthat$$$${\pi }^{ie}+\frac{1}{2}=0$$

Commented by mr W last updated on 17/Feb/20

$$certainly{it}^{\prime }snotyourfaultthatitis$$$$notequaltozero.itisso,nobodyis$$$$toblameforit.$$

Commented by mr W last updated on 17/Feb/20

$${\pi }^{ie}+\frac{1}{2}=e^{i\left(e\ln \pi \right)}+\frac{1}{2}=\cos \left(e\ln \pi \right)+\frac{1}{2}+i\sin \left(e\ln \pi \right)$$$$\approx -0.499553+0.029890i\ne 0$$

Commented by mr W last updated on 17/Feb/20

$$somethingiswrongwiththequestion.$$

Commented by M±th+et£s last updated on 17/Feb/20

$$thanksforallaboutthesolutiondandiam$$$$sosorryitsmyfaultit{doesn}^{\prime }tequalzero$$$$$$

Answered by MJS last updated on 17/Feb/20

$$a^{\mathrm{i}b}=$$$$[a>0]$$$$=\cos \left(b\ln a\right)+\mathrm{i}\sin \left(b\ln a\right)$$$$\{\begin{array}{l}\frac{1}{2}+\cos \left(b\ln a\right)=0\\ \sin \left(b\ln a\right)=0\end{array}$$$$\{\begin{array}{l}b\ln a=2k_1\pi \pm \frac{2\pi }{3};k_1\in \mathbb{Z}\\ b\ln a=k_2\pi ;k_2\in \mathbb{Z}\end{array}$$$$\Rightarrow k_2=2k_1\pm \frac{2}{3}\Rightarrow k_1\in \mathbb{Z}\mathrm{xor}{k}_2\in \mathbb{Z}\Rightarrow \mathrm{no}\mathrm{solution}$$$$\Rightarrow \mathrm{\forall }a\in {\mathbb{R}}^{+}\mathrm{\forall }b\in \mathbb{R}:a^{\mathrm{i}b}+\frac{1}{2}\ne 0$$$$$$$$z=\cos \left(b\ln a\right)+\mathrm{i}\sin \left(b\ln a\right)$$$$\Rightarrow \mathrm{abs}z=1\wedge \arg z=b\ln a$$$$\Rightarrow {\pi }^{\mathrm{ie}}={\mathrm{e}}^{\mathrm{ie}\ln \pi }\approx -.999553+.0298898\mathrm{i}$$$$\Rightarrow {\pi }^{\mathrm{ie}}+\frac{1}{2}\ne 0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com