Question-7582

Question Number 7582 by A WLAN last updated on 04/Sep/16

Commented by A WLAN last updated on 04/Sep/16

$${How}\:{to}\:{solve}\:{this}\:{problem}? \\ $$

Answered by Yozzia last updated on 04/Sep/16

w=1/z. You can plot the initial point z and the point w on an Argand diagram to deduce the angle of rotation. z=1⇒w=1/1=z⇒ angle of rotation is 0° about the origin. z=1 is invariant under the transformation. z=i⇒w=1/i=−i=−z⇒ angle of rotation is 180° about the origin. −−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Alternatively, we can find the argument of (w/z). Let w=r_1 e^(iφ ) and z=r_2 e^(iθ) . (w/z)=(r_1 /r_2 )e^(i(φ−θ)) ⇒arg(w/z)=φ−θ=arg(w)−arg(z) For z=1⇒arg(z)=θ=tan^(−1) (0/1)=0 radians ∵ w=1/z⇒ w=1⇒arg(w)=φ=tan^(−1) (0/1)=0 radians ∴ arg(w/z)=0−0=0 radians. ⇒The angle between w and z is 0 radians. ⇒The angle of rotation about the origin is 0 radians. z=i⇒ arg(z)=θ=lim_(x→0^+ ) tan^(−1) (1/x)=(π/2) w=1/z=1/i=−i⇒ arg(w)=φ=lim_(x→0^+ ) tan((−1)/x)=−lim_(x→0^+ ) tan^(−1) (1/x)=((−π)/2) ∴ arg(w/z)=((−π)/2)−(π/2)=−π. ⇒ The angle between w and z is π radians. ⇒The angle of rotation is π radians about the origin.

$${w}=\mathrm{1}/{z}.\:{You}\:{can}\:{plot}\:{the}\:{initial}\:{point}\:{z} \\ $$$${and}\:{the}\:{point}\:{w}\:{on}\:{an}\:{Argand}\:{diagram}\:{to}\:{deduce}\:{the}\:{angle}\:{of} \\ $$$${rotation}. \\ $$$${z}=\mathrm{1}\Rightarrow{w}=\mathrm{1}/\mathrm{1}={z}\Rightarrow\:{angle}\:{of}\:{rotation}\:{is}\:\mathrm{0}°\: \\ $$$${about}\:{the}\:{origin}.\:{z}=\mathrm{1}\:{is}\:{invariant}\:{under}\:{the}\:{transformation}. \\ $$$${z}={i}\Rightarrow{w}=\mathrm{1}/{i}=−{i}=−{z}\Rightarrow\:{angle}\:{of}\:{rotation}\:{is}\:\mathrm{180}°\: \\ $$$${about}\:{the}\:{origin}.\: \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Alternatively},\:{we}\:{can}\:{find}\:{the}\:{argument} \\ $$$${of}\:\frac{{w}}{{z}}.\:{Let}\:{w}={r}_{\mathrm{1}} {e}^{{i}\phi\:} {and}\:{z}={r}_{\mathrm{2}} {e}^{{i}\theta} . \\ $$$$\frac{{w}}{{z}}=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }{e}^{{i}\left(\phi−\theta\right)} \Rightarrow{arg}\left({w}/{z}\right)=\phi−\theta={arg}\left({w}\right)−{arg}\left({z}\right) \\ $$$${For}\:{z}=\mathrm{1}\Rightarrow{arg}\left({z}\right)=\theta={tan}^{−\mathrm{1}} \frac{\mathrm{0}}{\mathrm{1}}=\mathrm{0}\:{radians} \\ $$$$\because\:{w}=\mathrm{1}/{z}\Rightarrow\:{w}=\mathrm{1}\Rightarrow{arg}\left({w}\right)=\phi={tan}^{−\mathrm{1}} \frac{\mathrm{0}}{\mathrm{1}}=\mathrm{0}\:{radians} \\ $$$$\therefore\:{arg}\left({w}/{z}\right)=\mathrm{0}−\mathrm{0}=\mathrm{0}\:{radians}.\: \\ $$$$\Rightarrow{The}\:{angle}\:{between}\:{w}\:{and}\:{z}\:{is}\:\mathrm{0}\:{radians}. \\ $$$$\Rightarrow{The}\:{angle}\:{of}\:{rotation}\:{about}\:{the}\:{origin}\:{is}\:\mathrm{0}\:{radians}. \\ $$$${z}={i}\Rightarrow\:{arg}\left({z}\right)=\theta=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}=\frac{\pi}{\mathrm{2}} \\ $$$${w}=\mathrm{1}/{z}=\mathrm{1}/{i}=−{i}\Rightarrow\:{arg}\left({w}\right)=\phi=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{tan}\frac{−\mathrm{1}}{{x}}=−\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}=\frac{−\pi}{\mathrm{2}} \\ $$$$\therefore\:{arg}\left({w}/{z}\right)=\frac{−\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}=−\pi. \\ $$$$\Rightarrow\:{The}\:{angle}\:{between}\:{w}\:{and}\:{z}\:{is}\:\pi\:{radians}. \\ $$$$\Rightarrow{The}\:{angle}\:{of}\:{rotation}\:{is}\:\pi\:{radians}\:{about}\:{the}\:{origin}. \\ $$$$ \\ $$

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