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

  • MHB
  • Thread starter araz1
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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.
  • #1
araz1
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0
Hi
Show that z-(1/z)=i2rsinx, given z=r(cosx+isinx)
Thanks.
 
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  • #2
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 $\bar z = r(\cos x-i\sin x)$.
 
  • #3
No it is not z minus its conjugate. it is z minus (1 over z).
 
  • #4
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 $2-\frac 12=\frac 32\ne i\cdot 2\cdot \sin 0=0$.
 
  • #5
Your example makes sense, but that is a question in a book. Does it matter that z is in polar form?
 
  • #6
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.
 
  • #7
Yes there must be an error. Thank you for that.
 

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

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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