Question

Find the quotient of the complex numbers. Express your answer in trigonometric form.
z1= 5(Cos25degrees) + i sin(25degrees))
z2= 2(cos80degrees) + isin(80 degrees))

A. 10(cos(105degrees) + isin(105degrees))
B. 5/2 (cos(305degrees) + isin(305degrees))
C. 3(cos(-55degrees) + i sin(-55 degrees))
D. 7(cos(105degrees)+ i sin(105degrees))

Answer 1

@hartnn

Answer 2

@hartnn

Answer 3

con you convert it in polar form ?

Answer 4

The numbers are already in polar form, (unless you mean writing them with e is polar form).
In exponential notation they become:\[\frac{ z_1 }{ z_2 }=\frac{ 5e ^{i25⁰} }{ 2e ^{i80⁰} }=\frac{ 5 }{ 2 }e ^{i(25⁰-80⁰)}=\frac{ 5 }{ 2 }e ^{-55⁰i}\]Now, if you convert back to polar (trigonometric) notation, it will seem that this answer's not listed. Just think about the angle of -55⁰...

Answer 5

express in trigometric is what it said

Answer 6

Well, if you convert back my answer, you get:\[\frac{ 5 }{ 2 }e ^{-55⁰}=\frac{ 5 }{ 2 }(\cos(-55⁰) + i \sin(-55⁰)) \]Now, adding 360⁰ doesn't make any difference, so answer B. is correct.

Answer 7

Do you know the notation: \[z = R e ^{i \theta}=R(\cos \theta + i \sin \theta)\]It works great, because you can use the simple rules for working with powers in all kinds of calculations with complex numbers. That is what I did!

Answer 8

well do you know what the conjugate of 1 - 3i is?

Answer 9

The conjugate of z = a + bi is defined as a - bi, so if you add them up, the complex parts cancel out...

Answer 10

im not sure what you mean

Answer 11

For this problem, you don't need fancy notations.
if z = a + bi, the a - bi is it's conjugate.
I.e. 2 + 5i and 2 - 5i are conjugates, also -4 - 25i and -4 + 25i.
So the conjugate of 1 - 3i is...?

Answer 12

1+3i?

Answer 13

Right!