How to represent this complex number?

In summary: Z = 1^(1/6) is Z = 1*e^(i*0) (which has no imaginary part), there are other solutions to the equation, such as Z = 1*e^(i*π/3) (which does have an imaginary part). However, since we are only concerned with the values of Z that fall within the given bounds (0 < θ < π/2), we can only consider the latter solution, which gives us an argument of π/3. Adding 2π to this value does not change the fact that it is still the only solution that satisfies both the equation and the given bounds. So, in summary, adding 2π allows us to find the appropriate
  • #1
Elvis 123456789
158
6

Homework Statement


Finding "polar" and "rectangular" representation of a complex number?

Make a table with three columns. Each row will contain three representations of a
complex number z: the “rectangular” expression z = a + bi (with a and b real); the “polar”
expression |z|, Arg(z); and a little picture of the complex plane with the complex number
marked on it. There are five rows, containing, in one column or another, the following
complex numbers:


(iv) A sixth root of 1 with argument θ such that 0 < θ < π/2

Homework Equations


I don't know if de moivre's theorem for Nth roots of complex numbers is relevant here

The Attempt at a Solution


I know that the modulus of Z = r = 1^(1/6) = 1 but I don't know how to find the Argument of Z

Z = 1^(1/6) e^(iθ)
 
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  • #2
Hint: ##(re^{i\theta})^6 = r^6e^{i6\theta}##
 
  • #3
I don't know anything about this, but now someone has answered, I'll tell you my way.
Start with the third column! Sketch the diagram and use trig to work out the other two columns.
 
  • #4
Elvis 123456789 said:

Homework Statement


Finding "polar" and "rectangular" representation of a complex number?

Make a table with three columns. Each row will contain three representations of a
complex number z: the “rectangular” expression z = a + bi (with a and b real); the “polar”
expression |z|, Arg(z); and a little picture of the complex plane with the complex number
marked on it. There are five rows, containing, in one column or another, the following
complex numbers:


(iv) A sixth root of 1 with argument θ such that 0 < θ < π/2

Homework Equations


I don't know if de moivre's theorem for Nth roots of complex numbers is relevant here

The Attempt at a Solution


I know that the modulus of Z = r = 1^(1/6) = 1 but I don't know how to find the Argument of Z

Z = 1^(1/6) e^(iθ)
If ##\ z=\sqrt[6]1 \,,\ ## then ##\ z^6 = 1\ .\ ## Right?

What is the argument of 1 ?
 
  • #5
SammyS said:
If ##\ z=\sqrt[6]1 \,,\ ## then ##\ z^6 = 1\ .\ ## Right?

What is the argument of 1 ?
wouldn't the argument of 1 just be zero since it doesn't have an imaginary part?
 
  • #6
Elvis 123456789 said:
wouldn't the argument of 1 just be zero since it doesn't have an imaginary part?
That's one answer. What about the others?
 
  • #7
Isn't that the
Merlin3189 said:
That's one answer. What about the others?
Isn't that the only choice since 0 < θ < pi/2?
 
  • #8
SammyS said:
If ##\ z=\sqrt[6]1 \,,\ ## then ##\ z^6 = 1\ .\ ## Right?

What is the argument of 1 ?
Since you say ##\ z^6 = 1\ .\ ## presumably ##6\times Arg(z)\ =\ Arg(1)## or ##6\times θ\ =\ Arg(1)##
That does not mean Arg(1) must be between 0 and π/2

if you take Arg(1) as 0, then θ = 0/6 = 0. But you are asked for θ >0

Edit: If 0 < θ < π/2 then ##6\times θ\ =\ Arg(1)## might mean 0 < Arg(1) < 6π/2
 
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  • #9
Merlin3189 said:
Since you say ##\ z^6 = 1\ .\ ## presumably ##6\times Arg(z)\ =\ Arg(1)## or ##6\times θ\ =\ Arg(1)##
That does not mean Arg(1) must be between 0 and π/2

if you take Arg(1) as 0, then θ = 0/6 = 0. But you are asked for θ >0
I already know the answer, which is

1^(1/6) = 1*e^(i*pi/3) = 1/2 +i*√3/2. So Arg(z) = pi/3

but I just can't see how to get there :/
 
  • #10
Mind if I hop in? You have that [itex] Z = 1^{1/6} [/itex]. 1 is a complex number so it can be represented in the form [itex] Z = (re^{i\theta})^{1/6} [/itex]. As you said 1 has no imaginary part so r = 1. Now you have [itex] Z = e^{\frac{i\theta}{6}} [/itex]. Notice that the argument is effectively reduced by a factor 6,So there are other values for the argument within your bounds. You had 0 as one solution for theta, in trigonometry, how would you shift the argument of a function and still retain the same answer? Hope this helps.
 
  • #11
bigguccisosa said:
Mind if I hop in? You have that [itex] Z = 1^{1/6} [/itex]. 1 is a complex number so it can be represented in the form [itex] Z = (re^{i\theta})^{1/6} [/itex]. As you said 1 has no imaginary part so r = 1. Now you have [itex] Z = e^{\frac{i\theta}{6}} [/itex]. Notice that the argument is effectively reduced by a factor 6,So there are other values for the argument within your bounds. You had 0 as one solution for theta, in trigonometry, how would you shift the argument of a function and still retain the same answer? Hope this helps.
Are you referring to the fact that you can add 2pi to the argument and have the same answer?
 
  • #12
Elvis 123456789 said:
Are you referring to the fact that you can add 2pi to the argument and have the same answer?
Yes
 
  • #13
You could also ask yourself what angle could I add up 6 times to get ##2\pi##?
 
  • #14
bigguccisosa said:
Yes
So if r = 1 then Z = e^(iθ/6) and 0 < θ/6 <pi/2, and since Z= 1^1/6 the complex number has no imaginary part. Therefore the argument of Z has to be either 0/6 or 2pi/6?

but since we have the restriction 0 < θ/6 < pi/2 , it must be 2pi/6 = pi/3?
 
  • #15
Elvis 123456789 said:
but since we have the restriction 0 < θ/6 < pi/2 , it must be 2pi/6 = pi/3?
That looks good to me.
Elvis 123456789 said:
I already know the answer, which is

1^(1/6) = 1*e^(i*pi/3) = 1/2 +i*√3/2. So Arg(z) = pi/3

but I just can't see how to get there :/
 
  • #16
bigguccisosa said:
That looks good to me.
One more question though, since Z = 1^(1/6), we said that the complex number had no imaginary part. This is true for argument of Z at θ = 0, but not at θ =pi/3. So how can we maintain equivalent answers by adding 2pi when doing so gives us an imaginary number we didn't have to begin with?
 
  • #17
This is how I understand it: while it is true that 1 has no imaginary part, it is still a complex number in the complex plane. It is equivalent to saying 1+ 0i. A complex number represented as a + bi. From this cartesian form you have that [itex] r = \sqrt{a^2 + b^2} [/itex]. Both our answers, when represented in cartesian form, will give the same value of r. This diagram from Wolfram Alpha may help.( Both answers give the same value of r, which is the radius of the circle in the complex plane). Hope this helps, perhaps someone else can explain it better than I have :smile:
 

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  • #18
bigguccisosa said:
This is how I understand it: while it is true that 1 has no imaginary part, it is still a complex number in the complex plane. It is equivalent to saying 1+ 0i. A complex number represented as a + bi. From this cartesian form you have that [itex] r = \sqrt{a^2 + b^2} [/itex]. Both our answers, when represented in cartesian form, will give the same value of r. This diagram from Wolfram Alpha may help.( Both answers give the same value of r, which is the radius of the circle in the complex plane). Hope this helps, perhaps someone else can explain it better than I have :smile:
Ok I think I understand it a little better now. Thank you all for your help!
 
  • #19
Agree with bigguccisosa.

Even ##\sqrt{1}## has the same issue. ##z^2\ =\ 1## so ##2\times Arg(z)\ =\ Arg(1)## and, at first, ##Arg(z)\ =\ 0##
Only when you let Arg(1) be 2π, do you get 2 x Arg(z) = 2π and Arg(z) = π giving you the second (real) sqrt of 1, which is -1

To get all 6 roots for ##\sqrt [6] {1}## you need to consider 6 values for Arg(1) suc as -6π to +6π (that's 7 actually: miss off either 6π or -6π, but not both) (or use 0 to 10π, but for me something involving 6 is more intuitive!)

But for me the easiest way is just to use the diagram that bigguccisosa showed to get the arguments of the roots. Here the unit circle divided into 6 is 60o or π/3 just by inspection (and giving roots at n x 60o. or nπ/3 )
And the a + ib coordinates are equally easy (for 3rd, 4th, 6th and 8th roots, because the tangents are well known fractions) as ## \pm \frac {1}{2} \ \pm \frac {\sqrt(3)}{2}i \ \ \ and \pm 1 + 0i##
 

1. What is a complex number?

A complex number is a number that contains both a real and imaginary component. It is written in the form a + bi, where a is the real part and bi is the imaginary part.

2. How do I represent a complex number on a graph?

A complex number can be represented on a graph using the complex plane. The real part is plotted on the horizontal axis and the imaginary part is plotted on the vertical axis. The point where the two axes intersect represents the complex number.

3. What is the difference between polar and rectangular form for representing a complex number?

Polar form represents a complex number in terms of its magnitude and angle, while rectangular form represents it in terms of its real and imaginary components. Polar form is written as r(cosθ + isinθ), while rectangular form is written as a + bi.

4. How do I convert a complex number from polar to rectangular form?

To convert a complex number from polar to rectangular form, you can use the formula a = rcosθ and b = rsinθ, where a is the real part, b is the imaginary part, r is the magnitude, and θ is the angle.

5. Can a complex number be represented as a vector?

Yes, a complex number can be represented as a vector in the complex plane. The length of the vector represents the magnitude of the complex number, and the direction of the vector represents the angle. This is similar to how we represent vectors in a Cartesian coordinate system.

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