How to convert this complex exponential to polar then scalar

Question

I have the equation:

$$0=L_1+L_2e^{j\theta_2}-L_3e^{j\theta_3}-L_4e^{j\theta_4}$$

And I want to show the steps to get a scalar equation for $\theta_3$ that I can put into my calculator and solve.

I have converted it to complex polar form:

$$0=L_1+L_2(cos\theta_2+jsin\theta_2)-L_3(cos\theta_3+jsin\theta_3)-L_4(cos\theta_4+jsin\theta_4)$$

And then grouped the real terms:

$$=L_1+L_2cos\theta_2-L_3cos\theta_3-L_4cos\theta_4$$

But I am confused where I put the imaginary terms now. Do they stay in the same equation, do I pull them out? If I leave them in the equation, how do I get an equation for $\theta_3$ without imaginary components?

Answer 1

Well, assuming that all the variables are real and $\text{j}^2=-1$. We can use Euler's formula:

$$e^{xi}=\cos\left(x\right)+\sin\left(x\right)i\tag1$$

So, we can set up a system of equations:

$$ \begin{cases} \Re\left(\text{L}_1+\text{L}_2\cdot e^{\theta_2\text{j}}-\text{L}_3\cdot e^{\theta_3\text{j}}-\text{L}_4\cdot e^{\theta_4\text{j}}\right)=0\\ \\ \Im\left(\text{L}_1+\text{L}_2\cdot e^{\theta_2\text{j}}-\text{L}_3\cdot e^{\theta_3\text{j}}-\text{L}_4\cdot e^{\theta_4\text{j}}\right)=0 \end{cases}\tag2 $$

Which gives:

$$ \begin{cases} \text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_3\cdot\cos\left(\theta_3\right)-\text{L}_4\cdot\cos\left(\theta_4\right)=0\\ \\ \text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_3\cdot\sin\left(\theta_3\right)-\text{L}_4\cdot\sin\left(\theta_4\right)=0 \end{cases}\tag3 $$

Set $\text{L}_3$ equal to each other:

$$\frac{\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_4\cdot\cos\left(\theta_4\right)}{\cos\left(\theta_3\right)}=\frac{\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_4\cdot\sin\left(\theta_4\right)}{\sin\left(\theta_3\right)}\tag4$$

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EDIT in order to see what I did to find equation $\left(4\right)$:

• First equation in the system $\left(3\right)$: $$\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_3\cdot\cos\left(\theta_3\right)-\text{L}_4\cdot\cos\left(\theta_4\right)=0\space\Longleftrightarrow\space$$ $$\text{L}_3=\frac{\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_4\cdot\cos\left(\theta_4\right)}{\cos\left(\theta_3\right)}$$
• Second equation in the system $\left(3\right)$: $$\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_3\cdot\sin\left(\theta_3\right)-\text{L}_4\cdot\sin\left(\theta_4\right)=0\space\Longleftrightarrow\space$$ $$\text{L}_3=\frac{\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_4\cdot\sin\left(\theta_4\right)}{\sin\left(\theta_3\right)}$$

Now, set them equal and you get my equation $\left(4\right)$

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So, we also get:

$$\tan\left(\theta_3\right)=\frac{\sin\left(\theta_3\right)}{\cos\left(\theta_3\right)}=\frac{\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_4\cdot\sin\left(\theta_4\right)}{\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_4\cdot\cos\left(\theta_4\right)}\tag5$$

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EDIT: When we want to solve $x$:

$$\tan\left(x\right)=\text{z}\space\Longleftrightarrow\space x=\text{n}\cdot\pi+\arctan\left(\text{z}\right)\tag6$$

Where $\text{n}\in\mathbb{Z}$ and $1+\text{z}^2\ne0$

So, for your problem:

$$\theta_3=\text{n}\cdot\pi+\arctan\left(\frac{\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_4\cdot\sin\left(\theta_4\right)}{\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_4\cdot\cos\left(\theta_4\right)}\right)\tag7$$

When:

$$1+\left(\frac{\text{L}_2\cdot\sin\left(\theta_2\right)-\text{L}_4\cdot\sin\left(\theta_4\right)}{\text{L}_1+\text{L}_2\cdot\cos\left(\theta_2\right)-\text{L}_4\cdot\cos\left(\theta_4\right)}\right)^2\ne0\tag8$$

Answer 2

I assume that the $L$ coefficients are real.

You can solve in the complex,

$$L_3e^{j\theta_3}=L_1+L_2e^{j\theta_2}-L_4e^{j\theta_4}$$

and by taking the arguments,

$$\theta_3=\arg(L_1+L_2e^{j\theta_2}-L_4e^{j\theta_4}).$$

By the conversion formulas between Cartesian and polar representations,

$$\theta_3=\arctan\frac{L_2\sin\theta_2-L_4\sin\theta_4}{L_1+L_2\cos\theta_2-L_4\cos\theta_4}+k\pi.$$