# Finding z-(1/z) Given z=r(cosx+isinx)

• MHB
• araz1
In summary, the conversation discusses the equation z-(1/z)=i2rsinx and confirms that it is not the same as z minus its conjugate. The notation for complex conjugates is also clarified. The example z=2 is used to show that the equation does not hold. It is noted that z can be in polar form without affecting the calculations. It is suggested that there may be a typo in the book.
araz1
Hi
Show that z-(1/z)=i2rsinx, given z=r(cosx+isinx)
Thanks.

That is not the case.
Did you perhaps mean $z-\bar z$?
The notation $\bar z$ is the so called complex conjugate of $z$, which is $\bar z = r(\cos x-i\sin x)$.

No it is not z minus its conjugate. it is z minus (1 over z).

araz said:
No it is not z minus its conjugate. it is z minus (1 over z).
Then it is not the case.
Consider for instance $z=2$. Then we get $2-\frac 12=\frac 32\ne i\cdot 2\cdot \sin 0=0$.

Your example makes sense, but that is a question in a book. Does it matter that z is in polar form?

No Polar form is just another way to write a complex number. It does not affect calculations.
Likely there is a typo in the book then.

Yes there must be an error. Thank you for that.

## 1. What is the formula for finding z-(1/z)?

The formula for finding z-(1/z) is z=r(cosx+isinx) - 1/z.

## 2. How do you find the value of z in the equation z-(1/z)?

To find the value of z, you need to know the values of r, cosx, and sinx, and then plug them into the formula z=r(cosx+isinx) - 1/z.

## 3. Can you explain the concept of z-(1/z)?

Z-(1/z) is a complex number that is obtained by subtracting the reciprocal of z from z. It is a common operation in complex number arithmetic and is used in various applications in mathematics and engineering.

## 4. What is the significance of finding z-(1/z)?

Finding z-(1/z) can help in simplifying complex expressions and solving equations involving complex numbers. It is also used in analyzing circuits, signal processing, and other mathematical and scientific fields.

## 5. Are there any special cases to consider when finding z-(1/z)?

Yes, there are a few special cases to consider, such as when z=0 or when the value of z is purely real or purely imaginary. In these cases, the formula for z-(1/z) may need to be modified or the result may be undefined.

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