Question #303054

z1=3.45∠980°

z2=-5+9i

  1. z1+z2 final answer in polar form
  2. z1-z2 final answer in trigonometric form
  3. z2*z1 final answer in exponential form
  4. z1/z2 final answer in rectangular form
1
Expert's answer
2022-03-02T08:04:13-0500
z1=3.45980°=3.45ei980°=3.45ei100°z_1=3.45\angle980\degree=3.45e^{i980\degree}=3.45e^{-i100\degree}

=3.45(cos(100°)+isin(100°))=3.45(\cos(-100\degree)+i\sin(-100\degree))

=3.45(cos(5π/9)isin(5π/9))=3.45(\cos(5\pi/9)-i\sin(5\pi/9))

=3.45cos(5π/9)3.45sin(5π/9)i=3.45\cos(5\pi/9)-3.45\sin(5\pi/9)i


z2=5+9i=106eitan1(9/5)z_2=-5+9i=\sqrt{106}e^{i\tan^{-1}(9/-5)}

=106eitan1(1.8)=\sqrt{106}e^{-i\tan^{-1}(1.8)}

=106(cos(tan1(1.8))isin(tan1(1.8)))=\sqrt{106}(\cos(\tan^{-1}(1.8))-i\sin(\tan^{-1}(1.8)))

=106cos(tan1(1.8)106sin(tan1(1.8))i=\sqrt{106}\cos(\tan^{-1}(1.8)-\sqrt{106}\sin(\tan^{-1}(1.8))i

1.


z1+z2=3.45cos(5π/9)3.45sin(5π/9)iz_1+z_2=3.45\cos(5\pi/9)-3.45\sin(5\pi/9)i

+(5+9i)=(3.45cos(5π/9)5)++(-5+9i)=(3.45\cos(5\pi/9)-5)+


+(3.45sin(5π/9)+9)i+(-3.45\sin(5\pi/9)+9)i

5.60+5.60i=5.602(cos(135°)+isin(135°))\approx-5.60+5.60i=5.60\sqrt{2}(\cos(135\degree)+i\sin(135\degree))

z1+z2=5.602(cos(135°)+isin(135°))z_1+z_2=5.60\sqrt{2}(\cos(135\degree)+i\sin(135\degree))

=5.602135°=5.602 cis(135°)=5.60\sqrt{2}\angle135\degree=5.60\sqrt{2}\ cis(135\degree)

2.


z1z2=3.45cos(5π/9)3.45sin(5π/9)iz_1-z_2=3.45\cos(5\pi/9)-3.45\sin(5\pi/9)i

(5+9i)=(3.45cos(5π/9)+5)+-(-5+9i)=(3.45\cos(5\pi/9)+5)+


+(3.45sin(5π/9)9)i+(-3.45\sin(5\pi/9)-9)i

4.4012.40i=13.16(cos(70.5°)+isin(70.5°))\approx4.40-12.40i=13.16(\cos(-70.5\degree)+i\sin(-70.5\degree))

z1z2=13.16(cos(70.5°)+isin(70.5°))z_1-z_2=13.16(\cos(-70.5\degree)+i\sin(-70.5\degree))

=13.162289.5°=13.162 cis(289.5°)=13.16\sqrt{2}\angle289.5\degree=13.16\sqrt{2}\ cis(289.5\degree)

3.


z1z2=3.45ei100°106eitan1(1.8)z_1\cdot z_2=3.45e^{-i100\degree}\cdot\sqrt{106}e^{-i\tan^{-1}(1.8)}

=3.45106ei(100+tan1(1.8))=3.45\sqrt{106}e^{-i(100+\tan^{-1}(1.8))}


=35.52ei161°=35.52e^{-i161\degree}



4.


z1/z2=3.45ei100°/(106eitan1(1.8))z_1/z_2=3.45e^{-i100\degree}/(\sqrt{106}e^{-i\tan^{-1}(1.8)})

=3.45106ei(100+tan1(1.8))=3.45\sqrt{106}e^{i(-100+\tan^{-1}(1.8))}


=0.3351ei39.0546°=0.3351e^{-i39.0546\degree}

=0.260.21i=0.26-0.21i


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