2021-11-30

Question
For a linear transformation $w=\left(a+jb\right)z$, the point $A=\left(1+j\right)$ in the z-plane was transformed to $A=\left(-5+14j\right)$ in the w-plane. Calculate $\left(a+jb\right)$, the magnification, and the rotation of the transformation.

### Answer & Explanation

Rex Gibbons

Step 1
Rotalion and magnition: - A transformation of the form $\omega =\alpha z,\alpha \cdot i$ complex no.
and if $\alpha =\lambda {e}^{i\theta }$, then $\lambda$ is magnition whine $\theta$ is Rotalion.
Now

so from
$-5+14j=\left(a+jb\right)\left(1+j\right)$
$-5+14j=\left(a-b\right)+j\left(a+b\right)$
$a-b=-5$
$a+b=14$
Solving
$a=\frac{9}{2}$
$b=\frac{19}{2}$
Hence
$\omega =\left(\frac{9}{2}+j\frac{19}{2}\right)z$
put $\frac{9}{2}=\lambda \mathrm{cos},\frac{19}{2}=\lambda \mathrm{sin}\theta$
$\lambda =\sqrt{\frac{221}{2}}$
$\mathrm{tan}=\frac{19}{9}$
$\theta ={\mathrm{tan}}^{-}1\left(\frac{19}{9}\right)$
$\because w=\sqrt{\frac{221}{2}}\cdot e\left\{{}^{{\mathrm{tan}}^{-1}\left(\frac{19}{9}\right)}\right\}\cdot z$
Now
(1) $a+j3=\frac{9}{2}+j\frac{19}{2}$
(2) $magnition=\sqrt{\frac{221}{2}}$
(3) $Rotelion={\mathrm{tan}}^{-1}\left(\frac{19}{9}\right)$

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