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Question Number 61210 by pooja24 last updated on 30/May/19

$$forwhatvalueof\theta ,e^{i\theta }=0$$

Commented by Tony Lin last updated on 30/May/19

$$e^{i\theta }=cos\theta +isin\theta$$$$cos\theta \in [-1,1]\in R$$$$sin\theta \in [-1,1]\in R$$$$\Rightarrow if{e}^{i\theta }=0,thensin\theta mustbe0$$$$\Rightarrow cos\theta mustbe0$$$$but{cos}^2\theta +{sin}^2\theta =1$$$$\Rightarrow e^{i\theta }\ne 0$$

Answered by MJS last updated on 30/May/19

$${\mathrm{e}}^{\mathrm{i}\theta }\mathrm{usually}\mathrm{is}\mathrm{defined}\mathrm{for}-\pi ⩽\theta <\pi$$$$r{\mathrm{e}}^{\mathrm{i}\theta }=r\cos \theta +\mathrm{i}r\sin \theta$$$$\mathrm{in}\mathrm{our}\mathrm{case}r=1\Rightarrow \mathrm{it}\mathrm{can}\mathrm{never}\mathrm{be}\mathrm{zero}\mathrm{because}$$$$\mathrm{depending}\mathrm{on}\theta \mathrm{it}\mathrm{stands}\mathrm{for}\mathrm{all}\mathrm{complex}$$$$\mathrm{numbers}a+b\mathrm{i}\mathrm{with}r=\sqrt{a^2+b^2}=1$$

Answered by aleks041103 last updated on 31/May/19

$$\theta \to i\mathrm{\infty }$$$$\Rightarrow e^{i\theta }=e^{-\mathrm{\infty }}=0$$

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