# Complex numbers

The phase of a complex number is z=re$$^{i\theta}$$

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

The "r" is the distance from the origin. Thus, $$z=re^{i\psi}$$
$$z= r (cos(\psi)+i sin (\psi) )$$.
Hence, for $$z=1+i$$, $$r= \sqrt{1^2 + 1^2}= \sqrt{2}$$.

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The "r" is the distance from the origin. Thus, $$z=re^{i\psi}$$
$$z= r (cos(\psi)+i sin (\psi) )$$.
Hence, for $$z=1+i$$, $$r= \sqrt{1^2 + 1^2}= \sqrt{2}$$.

If it was 1+4i

Would it be:
$$r= \sqrt{1^2 + 4^2}$$?

learningphysics
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If it was 1+4i

Would it be:
$$r= \sqrt{1^2 + 4^2}$$?

yeah, that's right.

Is a reasonable answer for this phase then:
/sqrt{2}e ^i(theta) ?

since its just 'i' and no angle is given

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?

"The phase of a complex number is z=re$$^{i\theta}$$"

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

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

"The phase of a complex number is z=re$$^{i\theta}$$"

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 ?

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.

HallsofIvy
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No, for any complex number, both r and $\theta$ are both given. Your original question was about r only, that's why nothing was said about $\theta$.

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

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

In your second example, 1+ 4i, $r= \sqrt{1^2+ 4^2}= \sqrt{17}$ and $\theta= arctan(4/1)= arectan(4)= 1.3 radians approximately. Am puzzled by your saying "The phase of a complex number is [itex]z=re^{i\theta}$. Normally the "phase" is given as an angle. I would have thought the "phase" of the number $x+ iy= r e^{i\theta}$ would be just $\theta$.

ahhh, 1 and 1 is a 45 degrees.

when looking at a more complicated one like sqrt[2e] ^ -i(pi)/4
is the entire exponent my imaginery part?
and the base is real?

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

learningphysics
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when looking at a more complicated one like sqrt[2e] ^ -i(pi)/4
is the entire exponent my imaginery part?
and the base is real?

The e is inside the square root?

hmm
the books a little iffy..

I'll say no its not actually.

They're very easy to visualize, it's just like a point in the plane described in polar coordinates.

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).

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

lol yeah but its sqrt[2] multiplied by e and raised to -i(pi)/4 ! :P

Am I correct in saying the exponent is imaginery?