Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

for what value of θ,  e^(iθ) =0

$${for}\:{what}\:{value}\:{of}\:\theta,\:\:{e}^{{i}\theta} =\mathrm{0}\:\: \\ $$

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

e^(iθ) =cosθ+isinθ       cosθ∈[−1,1]∈R       sinθ∈[−1,1]∈R  ⇒if e^(iθ) =0, then sinθ must be 0                                ⇒cosθ must be 0  but cos^2 θ+sin^2 θ=1  ⇒e^(iθ) ≠0

$${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$$$\:\:\:\:\:{cos}\theta\in\left[−\mathrm{1},\mathrm{1}\right]\in{R} \\ $$$$\:\:\:\:\:{sin}\theta\in\left[−\mathrm{1},\mathrm{1}\right]\in{R} \\ $$$$\Rightarrow{if}\:{e}^{{i}\theta} =\mathrm{0},\:{then}\:{sin}\theta\:{must}\:{be}\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{cos}\theta\:{must}\:{be}\:\mathrm{0} \\ $$$${but}\:{cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow{e}^{{i}\theta} \neq\mathrm{0} \\ $$

Answered by MJS last updated on 30/May/19

e^(iθ)  usually is defined for −π≤θ<π  re^(iθ) =rcos θ +irsin θ  in our case r=1 ⇒ it can never be zero because  depending on θ it stands for all complex  numbers a+bi with r=(√(a^2 +b^2 ))=1

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

Answered by aleks041103 last updated on 31/May/19

θ→i ∞  ⇒e^(iθ) =e^(−∞) =0

$$\theta\rightarrow{i}\:\infty \\ $$$$\Rightarrow{e}^{{i}\theta} ={e}^{−\infty} =\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com