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

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

$$whoknowswhatis2andwhatisz$$$$whenyouwritethemidentically?$$

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

$$\mathit{z}=\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }={\mathit{e}}^{\mathit{i}\mathit{\theta }}$$$$\mathit{m}\mathit{t}\mathit{q}\frac{1}{1+\mathit{z}}=\frac{1}{2}(1-\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\theta }/2\right))$$$$\mathit{o}\mathit{n}\mathit{a}\frac{1}{1+\mathit{z}}=\frac{1}{{\mathit{e}}^{\mathit{i}(\frac{\mathit{\theta }}{2}-\frac{\mathit{\theta }}{2})}+{\mathit{e}}^{\mathit{i}(\frac{\mathit{\theta }}{2}+\frac{\mathit{\theta }}{2})}}$$$$=\frac{{\mathit{e}}^{-\mathit{i}\frac{\mathit{\theta }}{2}}}{2\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\theta }}{2}\right)}=\frac{\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\theta }}{2}\right)-\mathit{i}\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\theta }}{2}\right)}{2\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\theta }}{2}\right)}$$$$=\frac{1}{2}(1-\mathit{i}\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\theta }}{2}\right))$$$$..............profcedricjunior...............$$

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

$$\frac{1}{1+z}=\frac{1+cos\theta -isin\theta }{{(1+cos\theta )}^2+{sin}^2\theta }=\frac{2{cos}^2\left(\frac{\theta }{2}\right)-2isin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}{4{cos}^2\left(\frac{\theta }{2}\right)}$$$$=\frac{1}{2}(1-itan\left(\frac{\theta }{2}\right))✓$$$$$$

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

$$\mathrm{thanks}.\mathrm{mm}1342$$

Commented by Spillover last updated on 10/Aug/24

$$good$$

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