Question-210456

Question Number 210456 by peter frank last updated on 09/Aug/24

Commented by mr W last updated on 09/Aug/24

who knows what is 2 and what is z when you write them identically?

$${who}\:{knows}\:{what}\:{is}\:\mathrm{2}\:{and}\:{what}\:{is}\:{z} \\ $$$${when}\:{you}\:{write}\:{them}\:{identically}? \\ $$

Commented by Spillover last updated on 09/Aug/24

Commented by peter frank last updated on 10/Aug/24

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{correcting}\:\mathrm{the}\:\mathrm{question} \\ $$

Commented by lepuissantcedricjunior last updated on 10/Aug/24

z=cos𝛉+isin𝛉=e^(i𝛉) mtq (1/(1+z))=(1/2)(1−tan(𝛉/2)) on a (1/(1+z))=(1/(e^(i((𝛉/2)−(𝛉/2))) +e^(i((𝛉/2)+(𝛉/2))) )) =(e^(−i(𝛉/2)) /(2cos((𝛉/2)) ))=((cos((𝛉/2))−isin((𝛉/2)))/(2cos((𝛉/2)))) =(1/2)(1−itan((𝛉/2))) ..............prof cedric junior...............

$$\boldsymbol{{z}}=\boldsymbol{{cos}\theta}+\boldsymbol{{isin}\theta}=\boldsymbol{{e}}^{\boldsymbol{{i}\theta}} \\ $$$$\boldsymbol{{mtq}}\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{z}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\boldsymbol{{tan}}\left(\boldsymbol{\theta}/\mathrm{2}\right)\right) \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{z}}}=\frac{\mathrm{1}}{\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}−\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)} +\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}+\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{e}}^{−\boldsymbol{{i}}\frac{\boldsymbol{\theta}}{\mathrm{2}}} }{\mathrm{2}\boldsymbol{{cos}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)\:}=\frac{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)−\boldsymbol{{isin}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)}{\mathrm{2}\boldsymbol{{cos}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\boldsymbol{{itan}}\left(\frac{\boldsymbol{\theta}}{\mathrm{2}}\right)\right) \\ $$$$…………..{prof}\:{cedric}\:{junior}…………… \\ $$

Commented by Spillover last updated on 10/Aug/24

$${good} \\ $$

Answered by Spillover last updated on 09/Aug/24

Commented by peter frank last updated on 10/Aug/24

$$\mathrm{thanks}\:\mathrm{spillover} \\ $$

Answered by mm1342 last updated on 10/Aug/24

(1/(1+z))=((1+cosθ−isinθ)/((1+cosθ)^2 +sin^2 θ))=((2cos^2 ((θ/2))−2isin((θ/2))cos((θ/2)))/(4cos^2 ((θ/2)))) =(1/2)(1−itan((θ/2))) ✓

$$\frac{\mathrm{1}}{\mathrm{1}+{z}}=\frac{\mathrm{1}+{cos}\theta−{isin}\theta}{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{itan}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\checkmark \\ $$$$ \\ $$

Commented by peter frank last updated on 10/Aug/24

$$\mathrm{thanks}.\mathrm{mm1342} \\ $$

Commented by Spillover last updated on 10/Aug/24

$${good} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply