Complex Numbers - Finding in Polar Form

Let $\displaystyle z = \displaystyle{\frac{1}{2}} - \displaystyle{\frac{\sqrt{3}}{2}}i$ and $\displaystyle w = \displaystyle{\frac{\sqrt{3}}{2}} + \displaystyle{\frac{1}{2}}i$. If $\displaystyle u = \displaystyle{\frac{1}{z}} + \displaystyle{\frac{1}{w}}$, find $u$ in polar form.

I've managed to subtitute z and w into the equation for u to find the correct |u| value, but I can't find arg(u).

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Fratricide** Let $\displaystyle z = \displaystyle{\frac{1}{2}} - \displaystyle{\frac{\sqrt{3}}{2}}i$ and $\displaystyle w = \displaystyle{\frac{\sqrt{3}}{2}} + \displaystyle{\frac{1}{2}}i$. If $\displaystyle u = \displaystyle{\frac{1}{z}} + \displaystyle{\frac{1}{w}}$, find $u$ in polar form.

I've managed to subtitute z and w into the equation for u to find the correct |u| value, but I can't find arg(u).

Suppose that $x\cdot y\ne 0 $ then

$Arg(x + yi) = \left\{ {\begin{array}{{lr}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. $

Re: Complex Numbers - Finding in Polar Form

To be single valued, -Π/2 < tan^{-1}x < Π/2, all x.

Given x+iy, tan^{-1}(y/x) gives the same value for x&y in 1st and 3rd quadrants as for x&y in 2nd and 4th quadrants. So you have to know where x&y actually are on complex plane to pick the right angle.

(explanation of rule in previous post.)

Re: Complex Numbers - Finding in Polar Form

Sorry for the late reply, I've had a busy week since posting this thread.

Yet again, the notation you've used is unfamiliar to me. We're yet to cover arctan or the large brackets. Is there another way to explain it?

Re: Complex Numbers - Finding in Polar Form

If z=a+bi, let z’=a-bi

Multiply 1/z by z’/z’ and 1/w by w’/w’ and collect real and imaginary parts into u=a+bi.

Then plot u on the complex plane. lul is the distance from origin to u, and theta is the angle from positive x axis to radius.

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Fratricide** Sorry for the late reply, I've had a busy week since posting this thread.

Yet again, the notation you've used is unfamiliar to me. We're yet to cover arctan or the large brackets. Is there another way to explain it?

.

I have told you this before, here it is again. For the life of me, I cannot understand what sort of course you are doing in which you are asked questions having been given not reals tools to use.

**For any complex number** $\displaystyle z = a + bi$ to give the polar form it is necessary to find the argument.

That is associated with the tangent function.

If $\displaystyle a=0$ then the argument is $\displaystyle \pm\dfrac{\pi}{2}$.

If $\displaystyle a\ne 0$ then the quantity $\displaystyle \left| {\ffrac{b}{a}} \right|$ is the tangent of the argument given the correct sign.

To find the correct sine, you need to see where the number is located on the graph.

YOU need to know the tangents of each of the following:

$\displaystyle 0,~\dfrac{\pi}{2}.~\pi,~\dfrac{3\pi}{2}~\dfrac{\pi }{6}~\dfrac{\pi}{3}~\dfrac{2\pi}{3}~\dfrac{5\pi}{6 }~\dfrac{7\pi}{6}$ etc.

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Fratricide** Yet again, the notation you've used is unfamiliar to me ... the large brackets. Is there another way to explain it?

The large brackets merely indicate that a function is being defined in pieces and that each line represents a different piece.

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**JeffM** The large brackets merely indicate that a function is being defined in pieces and that each line represents a different piece.

The real point is he does not know how to use the $arctan()$ function!

Re: Complex Numbers - Finding in Polar Form

Good grief. Talk about mountain out of a mole-hill. Forget arctan function. Convert (x,y) to polar coordinates (r,theta) without thinking about it. Just look at the diagram and use elementary trigonometry.

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Hartlw** Good grief. Talk about mountain out of a mole-hill. Forget arctan function. Convert (x,y) to polar coordinates (r,theta) without thinking about it. Just look at the diagram and use elementary trigonometry.

Once again you lack of depth is in evidence.

Tell me if $w=-5+3i$ then how " converting (x,y) to polar coordinates (r,theta) without thinking about it" can possibility give a accurate $\arg(z)~?$

You see I would then require the student to give exact representations of the the six sixth roots of $z$.

What are they?

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Plato** Once again you lack of depth is in evidence.

Tell me if $w=-5+3i$ then how " converting (x,y) to polar coordinates (r,theta) without thinking about it" can possibility give a accurate $\arg(z)~?$

You see I would then require the student to give exact representations of the the six sixth roots of $z$.

What are they?

Look at (x,y):

tanθ’=3/5. θ’=31deg.

θ= 180-31 = 149deg = 2.6rad ± 2Π, the ± 2Π is obvious. trivial, and superflouus.

r=5.83@2.6rad, or whichever way you wish to express a complex number in polar form .

Re: Complex Numbers - Finding in Polar Form

Quote:

Originally Posted by

**Hartlw** Look at (x,y):

tanθ’=3/5. θ’=31deg.

θ= 180-31 = 149deg = 2.6rad ± 2Π, the ± 2Π is obvious. trivial, and superflouus.

r=5.83@2.6rad, or whichever way you wish to express a complex number in polar form .

I was afraid you would think that is a correct answer.

That answer would receive zero marks in any class I had control over.

It is a gross approximation, useless in modern mathematics. (no one uses degrees)

Re: Complex Numbers - Finding in Polar Form

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Originally Posted by

**Plato** I was afraid you would think that is a correct answer.

That answer would receive zero marks in any class I had control over.

It is a gross approximation, useless in modern mathematics. (no one uses degrees)

Arrogant nonsense. I gave the rght answer, acceptable in any course in complex analysis, and certainly at the level of the OP.

I notice that you always rely on insults when someone else gives the right answer. I didn't insult you for your post #10.

Re: Complex Numbers - Finding in Polar Form

Quote:

Originally Posted by

**Plato** .

I have told you this before, here it is again. For the life of me, I cannot understand what sort of course you are doing in which you are asked questions having been given not reals tools to use.

**For any complex number** $\displaystyle z = a + bi$ to give the polar form it is necessary to find the argument.

That is associated with the tangent function.

If $\displaystyle a=0$ then the argument is $\displaystyle \pm\dfrac{\pi}{2}$.

If $\displaystyle a\ne 0$ then the quantity $\displaystyle \left| {\ffrac{b}{a}} \right|$ is the tangent of the argument given the correct sign.

To find the correct sine, you need to see where the number is located on the graph.

YOU need to know the tangents of each of the following:

$\displaystyle 0,~\dfrac{\pi}{2}.~\pi,~\dfrac{3\pi}{2}~\dfrac{\pi }{6}~\dfrac{\pi}{3}~\dfrac{2\pi}{3}~\dfrac{5\pi}{6 }~\dfrac{7\pi}{6}$ etc.

I understand all of this fine (although it's written differently to what I'm familiar with). My problem was that couldn't grasp your specific use of arctan and the large brackets, but I have since cleared up my unfamiliarity with such notation. As for the course I'm doing -- it's early days. I'm not sure what the *real tools* are, but we've covered (from the textbook): The complex number system, C; Argand diagrams; Operations using complex numbers; Division of complex numbers and the complex conjugate; Polar form (modulus and argument) of complex numbers; Multiplication and division in polar form. This has involved completing exercises from the textbook, as well as a range of extended response questions (such as this one). My teacher is not going through the textbook in the order of the chapters, so that is the most likely reason why such things as arctan (something covered in chapter 2; complex numbers is chapter 3 -- the first chapter we're doing) are unfamiliar to me. This is also the first time this teacher has taught this course.

I thank you all for your willingness to help me through these problems. When contact time with my teacher is limited, the feedback and assistance MHF provides is extremely valuable.

Re: Complex Numbers - Finding in Polar Form

Quote:

Originally Posted by

**Plato** Once again you lack of depth is in evidence.

Tell me if $w=-5+3i$ then how " converting (x,y) to polar coordinates (r,theta) without thinking about it" can possibility give a accurate $\arg(z)~?$

You see I would then require the student to give exact representations of the the six sixth roots of $z$.

What are they?

The sixth roots of -5+3i, without use of the arctan function which OP felt unconfortable with, are:

1.34@(2.11rad+n2n/6), n= 0,1…,5

Would you like more decimal places?