Find modulus and Arg of the complex number 1 / (1 + i tan(theta))

[Qwertyuiop[]]

Hi, I have to find the modulus and the arg of this complex number: [math]\frac{1}{1+i \ tan \theta }[/math]I started by writing this in the general/algebraic form.
After multiplying the numerator and denominator with conjugate and simplifying i get: [math]cos^2\theta \:-\:i\:sin\theta cos\theta[/math]Then [math]a=cos^2\theta \:\\ b=-sin\theta cos\theta[/math]So the modulus is [math]\sqrt{\left(cos^2\theta \right)^2+\left(cos\theta sin\theta \right)^2}[/math] which simplifies to [math]cos\:\theta[/math]
To find the argument of z, we learned this formula [imath]cos\:\theta \:=\:\frac{a}{\left|z\right|}\:;\:sin\:\theta \:=\:\frac{b}{\left|z\right|}\:\:[/imath]
using this i get : [imath]cos\:\theta \:=\:cos\:\theta \:\:\:\:and\:sin\:\theta \:=-\:sin\:\theta[/imath]
cos theta is positive and sin theta is negative therefore it must be in the 4 quadrant so i wrote arg(z) as an inequality [math]0\:\le arg\left(z\right)\le -\frac{\pi }{2}[/math]Is the answer for mod(z) and arg(z) correct?

 

[khansaheb]

Hi, I have to find the modulus and the arg of this complex number: [math]\frac{1}{1+i \ tan \theta }[/math]I started by writing this in the general/algebraic form.
After multiplying the numerator and denominator with conjugate and simplifying i get: [math]cos^2\theta \:-\:i\:sin\theta cos\theta[/math]Then [math]a=cos^2\theta \:\\ b=-sin\theta cos\theta[/math]So the modulus is [math]\sqrt{\left(cos^2\theta \right)^2+\left(cos\theta sin\theta \right)^2}[/math] which simplifies to [math]cos\:\theta[/math]
To find the argument of z, we learned this formula [imath]cos\:\theta \:=\:\frac{a}{\left|z\right|}\:;\:sin\:\theta \:=\:\frac{b}{\left|z\right|}\:\:[/imath]
using this i get : [imath]cos\:\theta \:=\:cos\:\theta \:\:\:\:and\:sin\:\theta \:=-\:sin\:\theta[/imath]
cos theta is positive and sin theta is negative therefore it must be in the 4 quadrant so i wrote arg(z) as an inequality [math]0\:\le arg\left(z\right)\le -\frac{\pi }{2}[/math]Is the answer for mod(z) and arg(z) correct?

Correct ......

Another way:

\(\displaystyle \frac{1}{1 + i*tan\theta}\)

=\(\displaystyle \frac{cos\theta}{cos\theta +i sin\theta}\)

=\(\displaystyle cos\theta * e^{i * (-\theta)}\) ..........extract modulus and argument as before

 

[topsquark]

Hi, I have to find the modulus and the arg of this complex number: [math]\frac{1}{1+i \ tan \theta }[/math]I started by writing this in the general/algebraic form.
After multiplying the numerator and denominator with conjugate and simplifying i get: [math]cos^2\theta \:-\:i\:sin\theta cos\theta[/math]Then [math]a=cos^2\theta \:\\ b=-sin\theta cos\theta[/math]So the modulus is [math]\sqrt{\left(cos^2\theta \right)^2+\left(cos\theta sin\theta \right)^2}[/math] which simplifies to [math]cos\:\theta[/math]
To find the argument of z, we learned this formula [imath]cos\:\theta \:=\:\frac{a}{\left|z\right|}\:;\:sin\:\theta \:=\:\frac{b}{\left|z\right|}\:\:[/imath]
using this i get : [imath]cos\:\theta \:=\:cos\:\theta \:\:\:\:and\:sin\:\theta \:=-\:sin\:\theta[/imath]
cos theta is positive and sin theta is negative therefore it must be in the 4 quadrant so i wrote arg(z) as an inequality [math]0\:\le arg\left(z\right)\le -\frac{\pi }{2}[/math]Is the answer for mod(z) and arg(z) correct?

Well, your modulus is correct, but your argument has a flaw: you used [imath]\theta[/imath] for the argument, and you already have a [imath]\theta[/imath] in your problem.

Let's say that we have
[imath]z = r e^{i \alpha}[/imath]

Then [imath]r = cos( \theta )[/imath] as you already calculated. To get [imath]\alpha[/imath]:
[imath]z = r e^{i \alpha} = r \, cos(\alpha) + i r \, sin(\alpha)[/imath]

So
[imath]cos(\alpha) = \Re \left ( \dfrac{z}{r} \right ) = \dfrac{cos^2(\theta)}{cos(\theta)} = cos(\theta)[/imath]

and
[imath]sin(\alpha) = \Im \left ( \dfrac{z}{r} \right ) = \dfrac{-sin(\theta) \, cos(\theta )}{cos(\theta)} = -sin(\theta)[/imath]

So, the argument of z, [imath]\alpha[/imath] is [imath]- \theta[/imath]. (Which quadrant this is in depends on [imath]\theta[/imath].)

Of course, we need to mention when we state these that [imath]cos(-\theta) \neq 0[/imath] to be clear, but z doesn't actually exist when this happens, anyway. But you should make sure that this is covered, in general.

-Dan

 

[Dr.Peterson]

Have you considered what happens when [imath]\cos(\theta)<0[/imath]? In that case, the modulus can't be [imath]\cos(\theta)[/imath], can it?

Or was there a restriction on [imath]\theta[/imath] that you didn't mention?

 

[Qwertyuiop[]]

Have you considered what happens when [imath]\cos(\theta)<0[/imath]? In that case, the modulus can't be [imath]\cos(\theta)[/imath], can it?

Or was there a restriction on [imath]\theta[/imath] that you didn't mention?

no, they don't mention any restrictions on theta, they just say that theta is a given angle.

"Have you considered what happens when cos⁡(�)<0cos(θ)<0? In that case, the modulus can't be cos⁡(�)cos(θ), can it?"- If cos theta is negative that would make the modulus negative which is not possible.

 

[Qwertyuiop[]]

Well, your modulus is correct, but your argument has a flaw: you used [imath]\theta[/imath] for the argument, and you already have a [imath]\theta[/imath] in your problem.

Let's say that we have
[imath]z = r e^{i \alpha}[/imath]

Then [imath]r = cos( \theta )[/imath] as you already calculated. To get [imath]\alpha[/imath]:
[imath]z = r e^{i \alpha} = r \, cos(\alpha) + i r \, sin(\alpha)[/imath]

So
[imath]cos(\alpha) = \Re \left ( \dfrac{z}{r} \right ) = \dfrac{cos^2(\theta)}{cos(\theta)} = cos(\theta)[/imath]

and
[imath]sin(\alpha) = \Im \left ( \dfrac{z}{r} \right ) = \dfrac{-sin(\theta) \, cos(\theta )}{cos(\theta)} = -sin(\theta)[/imath]

So, the argument of z, [imath]\alpha[/imath] is [imath]- \theta[/imath]. (Which quadrant this is in depends on [imath]\theta[/imath].)

Of course, we need to mention when we state these that [imath]cos(-\theta) \neq 0[/imath] to be clear, but z doesn't actually exist when this happens, anyway. But you should make sure that this is covered, in general.

-Dan

How do you get the arg(z) to be -theta with cos theta and -sin theta?

 

[Qwertyuiop[]]

Correct ......

Another way:

\(\displaystyle \frac{1}{1 + i*tan\theta}\)

=\(\displaystyle \frac{cos\theta}{cos\theta +i sin\theta}\)

=\(\displaystyle cos\theta * e^{i * (-\theta)}\) ..........extract modulus and argument as before

How do you get from first step to the second ?

 

[Dr.Peterson]

no, they don't mention any restrictions on theta, they just say that theta is a given angle.

"Have you considered what happens when cos(θ)<0? In that case, the modulus can't be cos(θ), can it?"- If cos theta is negative that would make the modulus negative which is not possible.

Of course, we know that every complex number has a modulus; so you can't stop there! You need to look for (a) a small error in your work, and (b) how you can correct it. One thing you could do is to pick a specific angle whose cosine is negative, find the value of your expression in that case, and compare its actual modulus and argument with what your answer to the problem gives. That will show you, at least, how big or how small the error is.

Don't just ask others to tell you; give it some thought, and tell us your ideas. Our goal here is to help you learn to think for yourself. (This goes for all your responses!)

 

[topsquark]

How do you get the arg(z) to be -theta with cos theta and -sin theta?

Well, if
[imath]cos(\alpha) = cos(\theta)[/imath]

then [imath]\alpha = \pm \theta[/imath].

What does the sine condition say?

-Dan

 

[khansaheb]

How do you get from first step to the second ?

Did you try to resolve your doubt - using pencil/paper?

 

[pka]

I have to find the modulus and the arg of this complex number: [math]\Large\bf{\frac{1}{1+i \ tan \theta }}[/math]So the modulus is [math]\sqrt{\left(cos^2\theta \right)^2+\left(cos\theta sin\theta \right)^2}[/math] which simplifies to [math]cos\:\theta[/math]To find the argument of z, we learned this formula [imath]cos\:\theta \:=\:\frac{a}{\left|z\right|}\:;\:sin\:\theta \:=\:\frac{b}{\left|z\right|}\:\:[/imath]
using this i get : [imath]cos\:\theta \:=\:cos\:\theta \:\:\:\:and\:sin\:\theta \:=-\:sin\:\theta[/imath]
cos theta is positive and sin theta is negative therefore it must be in the 4 quadrant so i wrote arg(z) as an inequality [math]0\:\le arg\left(z\right)\le -\frac{\pi }{2[/math]

Some of this is a repeat, If [imath]z=a+b\bf{i}[/imath] then [imath]\dfrac{1}{z}=\dfrac{\overline{~z~}}{|z|^2}=\dfrac{a-b\bf{i}}{a^2+b^2}[/imath] and [imath]|z|=\sqrt{a^2+b^2}[/imath]
Next about the argument depends upon the quadrant in which [imath]z[/imath] is found.
If [imath]a\cdot b=0[/imath] then [imath]\text{Arg}(z)\in\left\{0,-\pi,\pm\dfrac{\pi}{2}\right\}[/imath]
If [imath]a\cdot b\ne 0[/imath] then let [imath]T = \arctan \left( {\left| {\dfrac{b}{a}} \right|} \right)[/imath] then [imath]Arg(z) = \left\{ \begin{gathered} T,\;z \in I \\ \pi - T,\;z \in II \\ - \pi + T,\;z \in III \\ - T,\;z \in IV \end{gathered} \right.[/imath]
Thus, as written the [imath]\tan(\Theta)[/imath] makes little to no sense unless the [imath]\Theta[/imath] is a constant.