z1 =3.45∠980° = 3.45 ∠ 260o
= 3.45(cos 260o + j sin 260o )
= −0.6 − 3.4i
z2 =5+9i
=N o w , r = ( 5 ) 2 + ( 9 ) 2 = 106 Now, \displaystyle{r}=\sqrt{{{\left({5}\right)}^{2}+{\left({9}\right)}^{2}}}=\sqrt{{{106}}} N o w , r = ( 5 ) 2 + ( 9 ) 2 = 106
A n d , θ = arctan ( 9 5 ) And, \displaystyle\theta= \arctan{{\left(\frac{9}{{5}}\right)}} A n d , θ = arctan ( 5 9 ) = 610
z2 = 106 \sqrt{{{106}}} 106 (cos(610 )+jsin(610 ))
(1) z 1 +z 2 in polar form
z1 +z2 = (−0.6 − 3.4j) + (5+9j) = 4.4 + 5.6j
In polar form,
N o w , r = ( 4.4 ) 2 + ( 5.6 ) 2 = 7.12 Now, \displaystyle{r}=\sqrt{{{\left({4.4}\right)}^{2}+{\left({5.6}\right)}^{2}}}={7.12} N o w , r = ( 4.4 ) 2 + ( 5.6 ) 2 = 7.12
A n d , θ = arctan ( 5.60 4.4 ) And, \displaystyle\theta= \arctan{{\left(\frac{5.60}{{4.4}}\right)}} A n d , θ = arctan ( 4.4 5.60 )
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
N o w , r = ( − 5.6 ) 2 + ( − 12.4 ) 2 = 13.606 Now, \displaystyle{r}=\sqrt{{{\left({-5.6}\right)}^{2}+{\left({-12.4}\right)}^{2}}}={13.606} N o w , r = ( − 5.6 ) 2 + ( − 12.4 ) 2 = 13.606
A n d , θ = arctan ( − 12.4 − 5.6 ) And, \displaystyle\theta= \arctan{{\left(\frac{-12.4}{{-5.6}}\right)}} A n d , θ = arctan ( − 5.6 − 12.4 )
= 2460
Thus,
z1 - z2 = 13.606 (cos (2460 )+jsin(2460 )) [Answer]
(3) z 2 *z 1 exponential form
z1 =3.45∠980° = 3.45 e j ( 980 180 π ) 3.45e^{j(\frac{980}{180}\pi )} 3.45 e j ( 180 980 π ) = 3.45 e j ( 5.44 π ) 3.45e^{j(5.44\pi )} 3.45 e j ( 5.44 π )
z2 = 5 +9j = 106 e j ( 61 180 π ) = 106 e j ( 0.34 π ) \sqrt{{{106}}}e^{j(\frac{61}{180}\pi)} = \sqrt{{{106}}}e^{j(0.34\pi)} 106 e j ( 180 61 π ) = 106 e j ( 0.34 π )
hence,
z 2 *z 1 = 3.45 e j ( 5.44 π ) ∗ 106 e j ( 0.34 π ) 3.45e^{j(5.44\pi )}*\sqrt{{{106}}}e^{j(0.34\pi)} 3.45 e j ( 5.44 π ) ∗ 106 e j ( 0.34 π )
=3.45 ∗ 106 e j ( 5.44 π + 0.34 π ) ) 3.45*\sqrt{{{106}}}e^{j(5.44\pi +0.34\pi) )} 3.45 ∗ 106 e j ( 5.44 π + 0.34 π ))
= 35.52 e j ( 5.78 π ) 35.52e^{j(5.78\pi )} 35.52 e j ( 5.78 π ) [Answer]
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