Question

Find the quotient z1/z2 of the complex numbers. Leave answer in polar form.
z1 = 1/8 (cos 2pi/3 + i sin 2pi/3)
z2= 1/3 (cos pi/4 + i sin pi/4)

Answer 1

\(\large e^{i\theta} = \cos \theta + i \sin \theta\)

Answer 2

cos 2pi/3 + i sin 2pi/3 = \(\large e^{i\frac{2\pi}{3}}\)

Answer 3

convert both complex numbers to exponent form..
and do the usual simplifications..

Answer 4

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (\cos 2\pi /3 + i \sin 2 \pi /3)}{ 1/3 (\cos \pi /4 + i \sin \pi /4) }} \)

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (e^{ i2 \pi /3})}{ 1/3 (e^{i \pi /4}) }} \)

\(\large \mathbb{\frac{z1}{z2} = 3/8 (e^{ i2 \pi /3 - i\pi/4})} \)

Answer 5

still here ?

Answer 6

I'm here. Just processing lol

Answer 7

good, 3rd step above is just simple division of a fraction...

Answer 8

Sorry I'm super tired (acute insomniac)

Answer 9

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (\cos 2\pi /3 + i \sin 2 \pi /3)}{ 1/3 (\cos \pi /4 + i \sin \pi /4) }} \)

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (e^{ i2 \pi /3})}{ 1/3 (e^{i \pi /4}) }} \)

\(\large \mathbb{\frac{z1}{z2} = \frac{3}{8}(e^{ i2 \pi /3 - i\pi/4})} \)


\(\large \mathbb{\frac{z1}{z2} = \frac{3}{8} (e^{ i 5 \pi /12})} \)

Answer 10

change it back to polar form again

Answer 11

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (\cos 2\pi /3 + i \sin 2 \pi /3)}{ 1/3 (\cos \pi /4 + i \sin \pi /4) }} \)

\(\large \mathbb{\frac{z1}{z2} = \frac{ 1/8 (e^{ i2 \pi /3})}{ 1/3 (e^{i \pi /4}) }} \)

\(\large \mathbb{\frac{z1}{z2} = \frac{3}{8}(e^{ i2 \pi /3 - i\pi/4})} \)


\(\large \mathbb{\frac{z1}{z2} = \frac{3}{8} (e^{ i 5 \pi /12})} \)


\(\large \mathbb{\frac{z1}{z2} = \frac{3}{8} (\cos 5 \pi /12 + i \sin 5 \pi /12) } \)

Answer 12

see if that makes some sense..

Answer 13

Thank you so much

Answer 14

It's a little more clear now :)

Answer 15

np, glad to hear that :)