Answer to Question #303040 in Algebra for RAIN

Question

z₁=3.45∠980°

z₂=5+9i

1. z1+z2 in polar form
2. z1-z2 trigonometric form
3. z2*z1 exponential form

Answer 1

z₁=3.45∠980° = 3.45 ∠ 260^o

= 3.45(cos 260^o + j sin 260^o)

= −0.6 − 3.4i

z₂=5+9i

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

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

z₂ = $\sqrt{{{106}}}$ (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,

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

$$And, \displaystyle\theta= \arctan{{\left(\frac{5.60}{{4.4}}\right)}}$$

• =51.84^o

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

(2) z1-z2 trigonometric form

z₁- z₂ =( −0.6 − 3.4i) - (5+9i)

= -5.6 - 12.4j

In trigonometric form

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

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

= 246⁰

Thus,

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

(3) z₂*z₁ exponential form

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

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

hence,

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

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

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