Question #303040

z1=3.45∠980°

z2=5+9i


  1. z1+z2 in polar form
  2. z1-z2 trigonometric form
  3. z2*z1 exponential form
1
Expert's answer
2022-03-04T06:54:31-0500


z1=3.45∠980° = 3.45 ∠ 260o


= 3.45(cos 260o + j sin 260o)


= −0.6 − 3.4i

z2=5+9i

=Now,r=(5)2+(9)2=106Now, \displaystyle{r}=\sqrt{{{\left({5}\right)}^{2}+{\left({9}\right)}^{2}}}=\sqrt{{{106}}}

And,θ=arctan(95)And, \displaystyle\theta= \arctan{{\left(\frac{9}{{5}}\right)}} = 610

z2 = 106\sqrt{{{106}}} (cos(610)+jsin(610))


(1) z1+z2 in polar form


z1+z2 = (−0.6 − 3.4j) + (5+9j) = 4.4 + 5.6j


In polar form,



Now,r=(4.4)2+(5.6)2=7.12Now, \displaystyle{r}=\sqrt{{{\left({4.4}\right)}^{2}+{\left({5.6}\right)}^{2}}}={7.12}


And,θ=arctan(5.604.4)And, \displaystyle\theta= \arctan{{\left(\frac{5.60}{{4.4}}\right)}}
  • =51.84o


Hence, 4.40 + 5.60j = 7.12 ∠ 51.84o [Answer]


(2) z1-z2 trigonometric form


z1- z2 =( −0.6 − 3.4i) - (5+9i)

= -5.6 - 12.4j


In trigonometric form


Now,r=(5.6)2+(12.4)2=13.606Now, \displaystyle{r}=\sqrt{{{\left({-5.6}\right)}^{2}+{\left({-12.4}\right)}^{2}}}={13.606}

And,θ=arctan(12.45.6)And, \displaystyle\theta= \arctan{{\left(\frac{-12.4}{{-5.6}}\right)}}

= 2460


Thus,

z1- z2 = 13.606 (cos (2460)+jsin(2460)) [Answer]



(3) z2*z1 exponential form


z1=3.45∠980° = 3.45ej(980180π)3.45e^{j(\frac{980}{180}\pi )} = 3.45ej(5.44π)3.45e^{j(5.44\pi )}

z2 = 5 +9j = 106ej(61180π)=106ej(0.34π)\sqrt{{{106}}}e^{j(\frac{61}{180}\pi)} = \sqrt{{{106}}}e^{j(0.34\pi)}


hence,


z2*z1 = 3.45ej(5.44π)106ej(0.34π)3.45e^{j(5.44\pi )}*\sqrt{{{106}}}e^{j(0.34\pi)}

=3.45106ej(5.44π+0.34π))3.45*\sqrt{{{106}}}e^{j(5.44\pi +0.34\pi) )}


= 35.52ej(5.78π)35.52e^{j(5.78\pi )} [Answer]

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