How can you calculate the arctan of e.g. 2(sqrt3)/6 without a calculator?

𝜃 = arctan(2√3∕6) ⇒ tan 𝜃 = sin 𝜃∕cos 𝜃 = 2√3∕6 = √3∕3 = 1∕√3 So, we have the ratio sin 𝜃 : cos 𝜃 = 1 : √3, which means that if sin 𝜃 = 𝑎, then cos 𝜃 = 𝑎√3 From the Pythagorean identity we have sin²𝜃 + cos²𝜃 = 1 ⇒ 𝑎² + 3𝑎² = 1 ⇒ 𝑎 = ±1∕2 So, sin 𝜃 = ±1∕2, cos 𝜃 = ±√3∕2, which means that the angle we're looking for is either in the first quadrant (sin 𝜃, cos 𝜃 > 0) or the third quadrant (sin 𝜃, cos 𝜃 < 0). Either way we're dealing with a right triangle whose legs are of lengths 1∕2 and √3∕2 and whose hypotenuse is 1. By reflecting this triangle over its longer leg, we get an equilateral triangle (in this case all three sides are equal to 1), and thereby we know that the original triangle must be a 30-60-90 triangle. Now we can use soh-cah-toa to find out that sin 30° = 1∕2 and cos 30° = √3∕2 So, in the first quadrant 𝜃 = 30°, or in radians 𝜃 = 𝜋∕6, and in the third quadrant 𝜃 = 𝜋∕6 − 𝜋 = −5𝜋∕6

I've been trying to understand how to find ALL polar coordinates for a week and it's getting a little bit frustrating. Could someone explain to me, given polar coordinates of a point, how to find ALL the coordinates? The thing that doesn't make me understand is this 2nπ thing or adding the 2nπ. I just don't get the method....

If you have a solution at π/2, then you have solutions every 2π. π/2 + 2π is a solution, π/2 + 2π + 2π is a solution. Because rotating by 2π is a complete rotation around the circle. So if you have 3Θ = π/4 +2πn Θ=(1/3)(π/4 + 2πn) Θ= π/12 + (2/3)πn Then it is a matter of picking the right n to get the solution in the required range

How is A complex number question such as z=10 and theta = 210 degrees answered with a square root? it says finding a is: a = 10cos210degrees and the answer is: a=-5sqrt3 Well, why must the answer be -5sqrt3 and not -8.660254038 rounded to -8.660 We feel we are missing something somewhere to not know this. Help, please.

They using special triangles(check trig section) to determine the answer. It looks like they wanted the exact answer.

What is the use of the exponential form i didn't understand it ?

It's more compact and easier to write, but I'm guessing we'd have to get into Euler's sequences in calculus to get why, exactly. I don't understand, either.

For the exercise on polar and rectangular forms of complex numbers, can we use a calculator or is there a way to do the calculations by hand?

There is no simple way to evaluate trigonometric functions by hand for arbitrary angles. It is only feasible to use a calculator for most angles.

Can we take out polar form of pure imaginary numbers

Yes. The polar form of an imaginary number X×i equals Xcis(π/2).

do you use theta in degrees or radians?

All angles shall be in radians unless indication of a different unit is used. (For e^iθ, it is in radians.)

how do you go from 5(cos90+isin90) polar form to rectagular form?

cos90 = 0 sin90 = 1 so 5(cos90+isin90) = 5 ( 0 + 1*i ) = 5i as the rectangular form.

Main content

Precalculus

Course: Precalculus > Unit 3

Lesson 6: Polar form of complex numbers

Complex number forms review

Q: I've been trying to understand how to find ALL polar coordinates for a week and it's getting a little bit frustrating. Could someone explain to me, given polar coordinates of a point, how to find ALL the coordinates? The thing that doesn't make me understand is this 2nπ thing or adding the 2nπ. I just don't get the method....

If you have a solution at π/2, then you have solutions every 2π. π/2 + 2π is a solution, π/2 + 2π + 2π is a solution. Because rotating by 2π is a complete rotation around the circle. So if you have 3Θ = π/4 +2πn Θ=(1/3)(π/4 + 2πn) Θ= π/12 + (2/3)πn Then it is a matter of picking the right n to get the solution in the required range

Q: How is A complex number question such as z=10 and theta = 210 degrees answered with a square root? it says finding a is: a = 10cos210degrees and the answer is: a=-5sqrt3 Well, why must the answer be -5sqrt3 and not -8.660254038 rounded to -8.660 We feel we are missing something somewhere to not know this. Help, please.

They using special triangles(check trig section) to determine the answer. It looks like they wanted the exact answer.

Q: What is the use of the exponential form i didn't understand it ?

It's more compact and easier to write, but I'm guessing we'd have to get into Euler's sequences in calculus to get why, exactly. I don't understand, either.

Q: For the exercise on polar and rectangular forms of complex numbers, can we use a calculator or is there a way to do the calculations by hand?

There is no simple way to evaluate trigonometric functions by hand for arbitrary angles. It is only feasible to use a calculator for most angles.

Q: Can we take out polar form of pure imaginary numbers

Yes. The polar form of an imaginary number X×i equals Xcis(π/2).

Q: do you use theta in degrees or radians?

All angles shall be in radians unless indication of a different unit is used. (For e^iθ, it is in radians.)

Q: how do you go from 5(cos90+isin90) polar form to rectagular form?

cos90 = 0 sin90 = 1 so 5(cos90+isin90) = 5 ( 0 + 1*i ) = 5i as the rectangular form.

CCSS.Math:

HSN.CN.B.4

Google Classroom

Review the different ways in which we can represent complex numbers: rectangular, polar, and exponential forms.

What are the different complex number forms?


Rectangular		a, plus, b, i
Polar		r, left parenthesis, cosine, left parenthesis, theta, right parenthesis, plus, i, sine, left parenthesis, theta, right parenthesis, right parenthesis
Exponential		r, dot, e, start superscript, i, theta, end superscript

Rectangular form

start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, i

The rectangular form of a complex number is a sum of two terms: the number's start color #11accd, start text, r, e, a, l, end text, end color #11accd part and the number's start color #1fab54, start text, i, m, a, g, i, n, a, r, y, end text, end color #1fab54 part multiplied by i.

As such, it is really useful for adding and subtracting complex numbers.

We can also plot a complex number given in rectangular form in the complex plane. The real and imaginary parts determine the real and imaginary coordinates of the number.

Want to learn more about complex number rectangular form? Check out this video about the complex plane and this video about adding and subtracting complex numbers.

Polar form

Polar form emphasizes the graphical attributes of complex numbers: start color #e07d10, start text, a, b, s, o, l, u, t, e, space, v, a, l, u, e, end text, end color #e07d10 (the distance of the number from the origin in the complex plane) and start color #aa87ff, start text, a, n, g, l, e, end text, end color #aa87ff (the angle that the number forms with the positive Real axis). These are also called start color #e07d10, start text, m, o, d, u, l, u, s, end text, end color #e07d10 and start color #aa87ff, start text, a, r, g, u, m, e, n, t, end text, end color #aa87ff.

Note that if we expand the parentheses in the polar representation, we get the number's rectangular form:

start color #e07d10, r, end color #e07d10, left parenthesis, cosine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, plus, i, dot, sine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, right parenthesis, equals, start overbrace, start color #e07d10, r, end color #e07d10, cosine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, end overbrace, start superscript, start color #11accd, a, end color #11accd, end superscript, plus, start overbrace, start color #e07d10, r, end color #e07d10, sine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, end overbrace, start superscript, start color #1fab54, b, end color #1fab54, end superscript, dot, i

This form is really useful for multiplying and dividing complex numbers, because of their special behavior: the product of two numbers with absolute values start color #e07d10, r, start subscript, 1, end subscript, end color #e07d10 and start color #e07d10, r, start subscript, 2, end subscript, end color #e07d10 and angles start color #aa87ff, theta, start subscript, 1, end subscript, end color #aa87ff and start color #aa87ff, theta, start subscript, 2, end subscript, end color #aa87ff will have an absolute value start color #e07d10, r, start subscript, 1, end subscript, r, start subscript, 2, end subscript, end color #e07d10 and angle start color #aa87ff, theta, start subscript, 1, end subscript, plus, theta, start subscript, 2, end subscript, end color #aa87ff.

Want to learn more about complex number polar form? Check out this video.

Exponential form

start color #e07d10, r, end color #e07d10, dot, e, start superscript, i, start color #aa87ff, theta, end color #aa87ff, end superscript

Exponential form uses the same attributes as polar form, start color #e07d10, start text, a, b, s, o, l, u, t, e, space, v, a, l, u, e, end text, end color #e07d10 and start color #aa87ff, start text, a, n, g, l, e, end text, end color #aa87ff. It only displays them in a different way that is more compact. For example, the multiplicative property can now be written as follows:

left parenthesis, start color #e07d10, r, start subscript, 1, end subscript, end color #e07d10, dot, e, start superscript, i, start color #9d38bd, theta, start subscript, 1, end subscript, end color #9d38bd, end superscript, right parenthesis, dot, left parenthesis, start color #e07d10, r, start subscript, 2, end subscript, end color #e07d10, dot, e, start superscript, i, start color #9d38bd, theta, start subscript, 2, end subscript, end color #9d38bd, end superscript, right parenthesis, equals, start color #e07d10, r, start subscript, 1, end subscript, end color #e07d10, start color #e07d10, r, start subscript, 2, end subscript, end color #e07d10, dot, e, start superscript, i, left parenthesis, start color #9d38bd, theta, start subscript, 1, end subscript, plus, theta, start subscript, 2, end subscript, end color #9d38bd, right parenthesis, end superscript

This form stems from Euler's expansion of the exponential function e, start superscript, z, end superscript to any complex number z. The reasoning behind it is quite advanced, but its meaning is simple: for any real number x, we define e, start superscript, i, x, end superscript to be cosine, left parenthesis, x, right parenthesis, plus, i, sine, left parenthesis, x, right parenthesis.

Using this definition, we obtain the equivalence of exponential and polar forms:

start color #e07d10, r, end color #e07d10, dot, e, start superscript, i, start color #aa87ff, theta, end color #aa87ff, end superscript, equals, start color #e07d10, r, end color #e07d10, left parenthesis, cosine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, plus, i, sine, left parenthesis, start color #aa87ff, theta, end color #aa87ff, right parenthesis, right parenthesis

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Abby Hilsman
Posted 4 years ago. Direct link to Abby Hilsman's post “How can you calculate the...”
How can you calculate the arctan of e.g. 2(sqrt3)/6 without a calculator?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Jerry Nilsson
  Posted 4 years ago. Direct link to Jerry Nilsson's post “𝜃 = arctan(2√3∕6) ⇒ tan ...”
  𝜃 = arctan(2√3∕6) ⇒
  tan 𝜃 = sin 𝜃∕cos 𝜃 = 2√3∕6 = √3∕3 = 1∕√3
  
  So, we have the ratio
  sin 𝜃 : cos 𝜃 = 1 : √3, which means that if sin 𝜃 = 𝑎, then cos 𝜃 = 𝑎√3
  
  From the Pythagorean identity we have
  sin²𝜃 + cos²𝜃 = 1 ⇒ 𝑎² + 3𝑎² = 1 ⇒ 𝑎 = ±1∕2
  
  So,
  sin 𝜃 = ±1∕2, cos 𝜃 = ±√3∕2, which means that the angle we're looking for is either in the first quadrant (sin 𝜃, cos 𝜃 > 0) or the third quadrant (sin 𝜃, cos 𝜃 < 0).
  
  Either way we're dealing with a right triangle whose legs are of lengths
  1∕2 and √3∕2 and whose hypotenuse is 1.
  By reflecting this triangle over its longer leg, we get an equilateral triangle (in this case all three sides are equal to 1), and thereby we know that the original triangle must be a 30-60-90 triangle.
  
  Now we can use soh-cah-toa to find out that
  sin 30° = 1∕2 and cos 30° = √3∕2
  
  So, in the first quadrant 𝜃 = 30°, or in radians 𝜃 = 𝜋∕6,
  and in the third quadrant 𝜃 = 𝜋∕6 − 𝜋 = −5𝜋∕6
  Button navigates to signup page
  (21 votes)
alex.bertke18
Posted 7 years ago. Direct link to alex.bertke18's post “how do you go from 5(cos9...”
how do you go from 5(cos90+isin90) polar form to rectagular form?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Judith Gibson
  Posted 7 years ago. Direct link to Judith Gibson's post “cos90 = 0 sin90 = 1 so 5(...”
  cos90 = 0
  sin90 = 1
  so 5(cos90+isin90) = 5 ( 0 + 1*i ) = 5i as the rectangular form.
  Button navigates to signup page
  (18 votes)
Brie Fossling
Posted 6 years ago. Direct link to Brie Fossling's post “I've been trying to under...”
I've been trying to understand how to find ALL polar coordinates for a week and it's getting a little bit frustrating. Could someone explain to me, given polar coordinates of a point, how to find ALL the coordinates? The thing that doesn't make me understand is this 2nπ thing or adding the 2nπ. I just don't get the method....
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Kadmilos
  Posted 6 years ago. Direct link to Kadmilos's post “If you have a solution at...”
  If you have a solution at π/2, then you have solutions every 2π.
  π/2 + 2π is a solution, π/2 + 2π + 2π is a solution. Because rotating by 2π is a complete rotation around the circle.
  
  So if you have
  3Θ = π/4 +2πn
  Θ=(1/3)(π/4 + 2πn)
  Θ= π/12 + (2/3)πn
  
  Then it is a matter of picking the right n to get the solution in the required range
  Button navigates to signup page
  (6 votes)
smotyer
Posted 2 years ago. Direct link to smotyer's post “I think you should add on...”
I think you should add one more step between being at let's say: 8cos(120) and then all of a sudden saying well that's clearly -4sqrt3 or something. It's not explained in the video either. On my calculator it just tells me -6.928 and I don't know how to get the 'exact' answer. Can you add an extra step or just allow for an answer like -6.928 to also be considered correct? Thank you so much for this free resource. I don't want to seem ungrateful I just don't get how to do that part!
Button navigates to signup pageComment on smotyer's post “I think you should add on...”
(5 votes)
Answer
- mccoyziehl
  Posted 3 months ago. Direct link to mccoyziehl's post “cos(120 degrees) = cos(2p...”
  cos(120 degrees) = cos(2pi/3) = -sqrt(3)/2
  8*cos(120 degrees) = 8 * -sqrt(3)/2 = -4sqrt(3)
  
  (Exact trig values are found on the unit circle)
  Button navigates to signup page
  (1 vote)
Marlene Bond
Posted 3 years ago. Direct link to Marlene Bond's post “How is A complex number q...”
How is A complex number question such as z=10 and theta = 210 degrees answered with a square root? it says finding a is: a = 10cos210degrees

and the answer is:
a=-5sqrt3

Well, why must the answer be -5sqrt3 and not -8.660254038 rounded to -8.660

We feel we are missing something somewhere to not know this.

Help, please.
Button navigates to signup pageComment on Marlene Bond's post “How is A complex number q...”
(3 votes)
Answer
- cossine
  Posted 3 years ago. Direct link to cossine's post “They using special triang...”
  They using special triangles(check trig section) to determine the answer. It looks like they wanted the exact answer.
  Comment on cossine's post “They using special triang...”
  (4 votes)
Akshat
Posted 7 years ago. Direct link to Akshat's post “What is the use of the ex...”
What is the use of the exponential form i didn't understand it ?
Button navigates to signup pageComment on Akshat's post “What is the use of the ex...”
(2 votes)
Answer
- Mike Schmidt
  Posted 7 years ago. Direct link to Mike Schmidt's post “It's more compact and eas...”
  It's more compact and easier to write, but I'm guessing we'd have to get into Euler's sequences in calculus to get why, exactly. I don't understand, either.
  Comment on Mike Schmidt's post “It's more compact and eas...”
  (5 votes)
Joseph Arcila
Posted 3 years ago. Direct link to Joseph Arcila's post “Why the product of two nu...”
Why the product of two numbers with angles theta1 and theta2 will have an angle theta1+theta2 ?
and Why the argument of z^n = n*theta ?
Thanks!
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(4 votes)
Answer
Deniz Boloni-Turgut
Posted 6 years ago. Direct link to Deniz Boloni-Turgut's post “For the exercise on polar...”
For the exercise on polar and rectangular forms of complex numbers, can we use a calculator or is there a way to do the calculations by hand?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- andrewp18
  Posted 6 years ago. Direct link to andrewp18's post “There is no simple way to...”
  There is no simple way to evaluate trigonometric functions by hand for arbitrary angles. It is only feasible to use a calculator for most angles.
  Comment on andrewp18's post “There is no simple way to...”
  (4 votes)
Karan Mehta
Posted 5 years ago. Direct link to Karan Mehta's post “Can we take out polar for...”
Can we take out polar form of pure imaginary numbers
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(2 votes)
Answer
- AIR
  Posted 4 years ago. Direct link to AIR's post “Yes. The polar form of an...”
  Yes. The polar form of an imaginary number X×i equals Xcis(π/2).
  Comment on AIR's post “Yes. The polar form of an...”
  (3 votes)
Elliott Bork
Posted 4 years ago. Direct link to Elliott Bork's post “do you use theta in degre...”
do you use theta in degrees or radians?
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(2 votes)
Answer
- AIR
  Posted 3 years ago. Direct link to AIR's post “All angles shall be in ra...”
  All angles shall be in radians unless indication of a different unit is used. (For e^iθ, it is in radians.)
  Button navigates to signup page
  (2 votes)