# Can someone explain this complex math problem?

• Voltux
In summary, the 315 term in the polar coordinates is obtained by using the mathematical function (333*1000)/(sqrt(333^2+1000^2)) to get rid of the complex number in the denominator and then converting from rectangular to polar coordinates.
Voltux
I'm working out an impedance matching problem from a textbook (it is not part of any coursework) and I am trying to figure out how they get the 315 term in the polar coordinates below.

Z = (XC*RL)/(XC+RL)

= (-j333*(1000))/(-j333+1000)
= 315 , -71.58*
= 100 -j300 ohms

I calculated that atan(1000/-333) = 71.58* however I do not understand where they got the 315 from. I get 249 or 499 depending on positive or negative.

I understand that 315 would be the length of the ray, and -71.58 would be the angle as this is capacitive hence -j. e.g. converting from rectangular to polar if I understand correctly.

It was explained as (333*1000)/(sqrt(333^2+1000^2)) and that indeed does give us 315.943 ~316, however, I do not understand how they used the mathematical function to get to this point and I was hoping someone could explain what I'm missing.

Voltux said:
I'm working out an impedance matching problem from a textbook (it is not part of any coursework) and I am trying to figure out how they get the 315 term in the polar coordinates below.

Z = (XC*RL)/(XC+RL)

= (-j333*(1000))/(-j333+1000)
= 315 , -71.58*
= 100 -j300 ohms

I calculated that atan(1000/-333) = 71.58* however I do not understand where they got the 315 from. I get 249 or 499 depending on positive or negative.

I understand that 315 would be the length of the ray, and -71.58 would be the angle as this is capacitive hence -j. e.g. converting from rectangular to polar if I understand correctly.

It was explained as (333*1000)/(sqrt(333^2+1000^2)) and that indeed does give us 315.943 ~316, however, I do not understand how they used the mathematical function to get to this point and I was hoping someone could explain what I'm missing.
Getting rid of a complex number in the denominator of a fraction is done by multiplying the fraction by 1 in the form of the conjugate of the denominator over itself.

$$\frac {a + jb}{c + jd} = \frac {a + jb}{c + jd} \cdot \frac{c - jd}{c - jd} \\ = \frac{(a + jb)(c - jd)}{c^2 + d^2} = \frac{ac + bd }{c^2 + d^2} + \frac{j(bc - ad)}{c^2 + d^2}$$

Voltux

## 1. What is the best way to approach understanding a complex math problem?

The best way to approach understanding a complex math problem is to break it down into smaller, more manageable parts. Start by identifying the key components and concepts involved in the problem and then work through each part step-by-step.

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Improving your problem-solving skills for complex math problems takes practice and patience. Start by familiarizing yourself with different problem-solving strategies and techniques, and then apply them to different types of problems. Be sure to also seek help and guidance from others when needed.

## 3. Is it important to fully understand all the steps in solving a complex math problem?

Yes, it is important to fully understand all the steps in solving a complex math problem. This not only helps you to solve the current problem, but also builds a strong foundation for solving future problems. It also allows you to identify any mistakes or errors in your solution and correct them.

## 4. How can I check if my solution to a complex math problem is correct?

One way to check if your solution to a complex math problem is correct is to plug in your answer into the original problem and see if it satisfies all the given conditions. You can also try using a different method or approach to solve the problem and see if you get the same result.

## 5. Are there any resources or tools that can help me understand complex math problems?

Yes, there are many resources and tools available to help you understand complex math problems. Some examples include online tutorials, textbooks, study groups, and tutoring services. It can also be helpful to practice solving similar problems and seeking feedback from others.

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