Question #263934

Express the following in rectangular and polar form, if


Z1 = 3+ 4i


Z2= 2+3i


1. Z1*Z2


2. Z1-Z2


3. Z1/Z2


4. |Z1|


5. |Z2|



(2)if Z1=50<30°and Z2=30<60°find in rectangular


form the following


1. |Z1|


2. |Z2|


3. |Z1|-|Z2|


4. Z1*Z2


5. |Z2-Z1|


6. |Z2|/|Z1|

1
Expert's answer
2021-11-11T12:28:55-0500

(1)

1.

z1z2=(3+4i)(2+3i)=6+9i+8i12z_1\cdot z_2=(3+ 4i)(2+ 3i)=6+9i+8i-12

=6+17i=-6+17i

z1z2=(6)2+172=517|z_1\cdot z_2|=\sqrt{(-6)^2+17^2}=5\sqrt{17}

tanθ=176,90°<θ<180°\tan \theta=\dfrac{17}{-6}, 90\degree<\theta<180\degree

z1z2=517(180°tan1176)z_1\cdot z_2=5\sqrt{17}\angle\big(180\degree-\tan^{-1}\dfrac{17}{6}\big)

2.

z1z2=(3+4i)(2+3i)=1+iz_1-z_2=(3+ 4i)-(2+ 3i)=1+i

z1z2=12+12=2|z_1- z_2|=\sqrt{1^2+1^2}=\sqrt{2}

tanθ=11=1,0°<θ<90°\tan \theta=\dfrac{1}{1}=1, 0\degree<\theta<90\degree

z1z2=245°z_1-z_2=\sqrt{2}\angle45\degree

3.

z1/z2=(3+4i)/(2+3i)=(3+4i)(23i)22+32z_1/ z_2=(3+ 4i)/(2+ 3i)=\dfrac{(3+4i)(2-3i)}{2^2+3^2}

=69i+8i+1213=1813113i=\dfrac{6-9i+8i+12}{13}=\dfrac{18}{13}-\dfrac{1}{13}i

z1/z2=(1813)2+(113)2=51313|z_1/ z_2|=\sqrt{(\dfrac{18}{13})^2+(-\dfrac{1}{13})^2}=\dfrac{5\sqrt{13}}{13}

tanθ=1131813=118,270°<θ<360°\tan \theta=\dfrac{-\dfrac{1}{13}}{\dfrac{18}{13}}=-\dfrac{1}{18}, 270\degree<\theta<360\degree

z1/z2=51313(360°tan1118)z_1/ z_2=\dfrac{5\sqrt{13}}{13}\angle\big(360\degree-\tan^{-1}\dfrac{1}{18}\big)

4.


z1=32+42=5|z_1|=\sqrt{3^2+4^2}=5

z1=5+0i|z_1|=5+0i

z1=50°|z_1|=5\angle0\degree

5.


z2=22+32=13|z_2|=\sqrt{2^2+3^2}=\sqrt{13}

z2=13+0i|z_2|=\sqrt{13}+0i

z2=130°|z_2|=\sqrt{13}\angle0\degree

(2)

1.


z1=50+0i|z_1|=50+0i

2.


z2=30+0i|z_2|=30+0i



3.


z1z2=5030=20|z_1|-|z_2|=50-30=20


z1z2=20+0i|z_1|-|z_2|=20+0i



4.


z1z2=5030(cos(30°+60°)+isin(30°+60°))z_1\cdot z_2=50\cdot 30(\cos (30\degree+60\degree)+i\sin (30\degree+60\degree))




=1500(0+i(1))=1500(0+i(1))


z1z2=0+1500iz_1\cdot z_2=0+1500i

5.


z1=50(cos30°+isin30°)=253+25iz_1=50(\cos 30\degree+i\sin 30\degree)=25\sqrt{3}+25i

z2=30(cos60°+isin60°)=15+153iz_2=30(\cos 60\degree+i\sin 60\degree)=15+15\sqrt{3}i


z2z1=(15253)+(15325)iz_2-z_1=(15-25\sqrt{3})+(15\sqrt{3}-25)i

z2z1=(15253)2+(15325)2|z_2-z_1|=\sqrt{(15-25\sqrt{3})^2+(15\sqrt{3}-25)^2}

=2257503+1875+6757503+625=\sqrt{225-750\sqrt{3}+1875+675-750\sqrt{3}+625}

=1034153=10\sqrt{34-15\sqrt{3}}

z2z1=1034153+0i|z_2-z_1|=10\sqrt{34-15\sqrt{3}}+0i

6.


z2z1=3050=35\dfrac{|z_2|}{|z_1|}=\dfrac{30}{50}=\dfrac{3}{5}


z2/z1=35+0i|z_2|/|z_1|=\dfrac{3}{5}+0i


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