- #1

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This first example is a simple z=1+i, but where does the r come from for this?

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- Thread starter Oblio
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- #1

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This first example is a simple z=1+i, but where does the r come from for this?

- #2

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The "r" is the distance from the origin. Thus, [tex]z=re^{i\psi} [/tex]

[tex]z= r (cos(\psi)+i sin (\psi) ) [/tex].

Hence, for [tex]z=1+i[/tex], [tex] r= \sqrt{1^2 + 1^2}= \sqrt{2} [/tex].

[tex]z= r (cos(\psi)+i sin (\psi) ) [/tex].

Hence, for [tex]z=1+i[/tex], [tex] r= \sqrt{1^2 + 1^2}= \sqrt{2} [/tex].

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- #3

- #4

learningphysics

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If it was 1+4i

Would it be:

[tex] r= \sqrt{1^2 + 4^2} [/tex]?

yeah, that's right.

- #5

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Is a reasonable answer for this phase then:

/sqrt{2}e ^i(theta) ?

/sqrt{2}e ^i(theta) ?

- #6

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since its just 'i' and no angle is given

- #7

learningphysics

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Is a reasonable answer for this phase then:

/sqrt{2}e ^i(theta) ?

wait... phase usually refers to the angle... what exactly does the question ask you to find?

from your original post:

"The phase of a complex number is z=re[tex]^{i\theta}[/tex]"

that doesn't seem right... did you write this out word for word?

- #8

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wait... phase usually refers to the angle... what exactly does the question ask you to find?

from your original post:

"The phase of a complex number is z=re[tex]^{i\theta}[/tex]"

that doesn't seem right... did you write this out word for word?

No, phase is just the angle but there doesn't seem to be any angle so.. yeah...

exact words are: phase is the value of theta when z is expressed as z=re^i(theta).

Can you find an angle with just 1+i ?

- #9

learningphysics

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No, phase is just the angle but there doesn't seem to be any angle so.. yeah...

exact words are: phase is the value of theta when z is expressed as z=re^i(theta).

Can you find an angle with just 1+i ?

yes, you can. arctan(1/1). Draw the point on the complex plane... the x-coordinate is the real... y-coordinate is the complex part... what angle does the line from the origin to the point make with the positive x-axis...

That's the angle you need.

- #10

HallsofIvy

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In the "complex plane" or Argand diagram, any complex number x+ iy can be associated with the point (x,y). [itex]r (cos(\theta)+ i sin(\theta))= re^{i\theta}[/itex] is just that point given in polar coordinates. Since [itex]x^2+ y^2= r^2 cos^2(theta)+ r^2 sin^2(\theta)= r^2[/itex], [itex]r= \sqrt{x^2+ y^2}[/itex]. Since [itex]y/x= r^2 sin^2(\theta)/[r^2 cos^2(\theta)= tan(\theta)[/itex], [itex]\theta= arctan(y/x)[/itex].

In your first example, 1+i, [itex]r= \sqrt{1^2+ 1^2}= \sqrt{2}[/itex] and [itex]\theta= arctan(1/1)= arctan(1)= \pi/4[/itex].

In your second example, 1+ 4i, [itex]r= \sqrt{1^2+ 4^2}= \sqrt{17}[/itex] and [itex]\theta= arctan(4/1)= arectan(4)= 1.3 radians approximately.

Am puzzled by your saying "The phase of a complex number is [itex]z=re^{i\theta}[/itex]. Normally the "phase" is given as an angle. I would have thought the "phase" of the number [itex]x+ iy= r e^{i\theta}[/itex] would be just [itex]\theta[/itex].

- #11

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ahhh, 1 and 1 is a 45 degrees.

- #12

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is the entire exponent my imaginery part?

and the base is real?

- #13

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when theyre like this, all hope of visualizing it is gone.

- #14

learningphysics

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is the entire exponent my imaginery part?

and the base is real?

The e is inside the square root?

- #15

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hmm

the books a little iffy..

I'll say no its not actually.

the books a little iffy..

I'll say no its not actually.

- #16

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In your last example, sqrt(2) is the distance from the origin to the point in the plane and -pi/4 is the angle (in radians) starting from the real axis (or the x axis in the analogy).

- #17

learningphysics

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hmm

the books a little iffy..

I'll say no its not actually.

Then it shouldn't be hard to visualize... r = sqrt(2). theta = -pi/4

- #18

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Am I correct in saying the exponent is imaginery?

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