angle of rotation

logistic_guy

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Determine the angle of rotation at the point z0=2+i\displaystyle z_{0} = 2+i when w=z2\displaystyle w = z^2, and illustrate it for some particular curve. Show that the scale factor at that point is 25\displaystyle 2\sqrt{5}.
 
Determine the angle of rotation at the point z0=2+i\displaystyle z_{0} = 2+i when w=z2\displaystyle w = z^2, and illustrate it for some particular curve. Show that the scale factor at that point is 25\displaystyle 2\sqrt{5}.

show us your effort/s to solve this problem.
 
show us your effort/s to solve this problem.
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I will let f(z)=w=z2\displaystyle f(z) = w = z^2

then, the angle of rotation is:

θ=arg(dfdzz0)=arg(ddzz2z0)=arg(2zz0)=arg(2z0)=arg[ 2(2+i) ]\displaystyle \theta = \text{arg}\left(\frac{df}{dz}\bigg|_{z_0}\right) = \text{arg}\left(\frac{d}{dz}z^2\bigg|_{z_0}\right) = \text{arg}\left(2z\bigg|_{z_0}\right) = \text{arg}(2z_0) = \text{arg}[ \ 2(2 + i) \ ]

=tan124=tan1120.4636 rad26.6\displaystyle = \tan^{-1}\frac{2}{4} = \tan^{-1}\frac{1}{2} \approx 0.4636 \ \text{rad} \approx 26.6^{\circ}

And the factor is:

(dfdzz0)=42+22=16+4=20=25\displaystyle \left|\left(\frac{df}{dz}\bigg|_{z_0}\right)\right| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}

I will continue in the next post.
 
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