Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 152203 by peter frank last updated on 26/Aug/21

If x is real show that (2+i)^((1+3i)x) +(2−i)^((1−3i)x) is also real

$$\mathrm{If}\:\mathrm{x}\:\mathrm{is}\:\mathrm{real}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{2}+\mathrm{i}\right)^{\left(\mathrm{1}+\mathrm{3i}\right)\mathrm{x}} +\left(\mathrm{2}−\mathrm{i}\right)^{\left(\mathrm{1}−\mathrm{3i}\right)\mathrm{x}} \\ $$$$\mathrm{is}\:\mathrm{also}\:\mathrm{real} \\ $$

Commented by MJS_new last updated on 26/Aug/21

(e^(θi) r)^(a+bi) +(e^(−θi) r)^(a−bi) = =e^(−θb+θai) r^(a+bi) +e^(−θb−θai) r^(a−bi) = =(e^(θai) r^(bi) )(e^(−θb) r^a )+(e^(−θai) r^(−bi) )(e^(−θb) r^a ) easy to see the shape of these is e^(iφ) s+e^(−iφ) s=2scos φ which obviously is real

$$\left(\mathrm{e}^{\theta\mathrm{i}} {r}\right)^{{a}+{b}\mathrm{i}} +\left(\mathrm{e}^{−\theta\mathrm{i}} {r}\right)^{{a}−{b}\mathrm{i}} = \\ $$$$=\mathrm{e}^{−\theta{b}+\theta{a}\mathrm{i}} {r}^{{a}+{b}\mathrm{i}} +\mathrm{e}^{−\theta{b}−\theta{a}\mathrm{i}} {r}^{{a}−{b}\mathrm{i}} = \\ $$$$=\left(\mathrm{e}^{\theta{a}\mathrm{i}} {r}^{{b}\mathrm{i}} \right)\left(\mathrm{e}^{−\theta{b}} {r}^{{a}} \right)+\left(\mathrm{e}^{−\theta{a}\mathrm{i}} {r}^{−{b}\mathrm{i}} \right)\left(\mathrm{e}^{−\theta{b}} {r}^{{a}} \right) \\ $$$$\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is} \\ $$$$\mathrm{e}^{\mathrm{i}\phi} {s}+\mathrm{e}^{−\mathrm{i}\phi} {s}=\mathrm{2}{s}\mathrm{cos}\:\phi \\ $$$$\mathrm{which}\:\mathrm{obviously}\:\mathrm{is}\:\mathrm{real} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com