How to represent this complex number?

[Elvis 123456789]

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)

 

[LCKurtz]

Hint: $(re^{i\theta})^6 = r^6e^{i6\theta}$

 

[Merlin3189]

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.

 

[SammyS]

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 ?

 

[Elvis 123456789]

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?

 

[Merlin3189]

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?

 

[Elvis 123456789]

Isn't that the

That's one answer. What about the others?

Isn't that the only choice since 0 < θ < pi/2?

 

[Merlin3189]

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

 

[Elvis 123456789]

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 :/

 

[bigguccisosa]

Mind if I hop in? You have that Z = 1^{1/6}. 1 is a complex number so it can be represented in the form Z = (re^{i\theta})^{1/6}. As you said 1 has no imaginary part so r = 1. Now you have Z = e^{\frac{i\theta}{6}}. 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.

 

[Elvis 123456789]

Mind if I hop in? You have that Z = 1^{1/6}. 1 is a complex number so it can be represented in the form Z = (re^{i\theta})^{1/6}. As you said 1 has no imaginary part so r = 1. Now you have Z = e^{\frac{i\theta}{6}}. 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?

 

[bigguccisosa]

Are you referring to the fact that you can add 2pi to the argument and have the same answer?

Yes

 

[LCKurtz]

You could also ask yourself what angle could I add up 6 times to get $2\pi$?

 

[Elvis 123456789]

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?

 

[bigguccisosa]

but since we have the restriction 0 < θ/6 < pi/2 , it must be 2pi/6 = pi/3?

That looks good to me.

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 :/

 

[Elvis 123456789]

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?

 

[bigguccisosa]

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 r = \sqrt{a^2 + b^2}. 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

 

[Elvis 123456789]

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 r = \sqrt{a^2 + b^2}. 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

Ok I think I understand it a little better now. Thank you all for your help!

 

[Merlin3189]

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 60^o or π/3 just by inspection (and giving roots at n x 60^o. 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 $\pm \frac {1}{2} \ \pm \frac {\sqrt(3)}{2}i \ \ \ and \pm 1 + 0i$