# Thread: differentiation and complex numbers

1. ## differentiation and complex numbers

okay just 2 quick ones

differentiate w = 6t^(-3/2) - 4t^(3/2)

and given that z = 1 + j and w = 2-3j find

w/z

these are the last 2 questions im stuck with and any help would be wicked cheers!.

okay just 2 quick ones

differentiate w = 6t^(-3/2) - 4t^(3/2)

and given that z = 1 + j and w = 2-3j find

w/z

these are the last 2 questions im stuck with and any help would be wicked cheers!.
What is that supposed to mean?

You need to be more precise about what $\displaystyle w$ is. And what you mean by $\displaystyle w/z$.

given that z = 1 + j and w = 2-3j find w/z
This complex division: $\displaystyle \frac{w}{z} = \frac{{w\overline z }}{{\left| z \right|^2 }}$

4. ## complex number

yes its a complex number division the equation i have is wz*/|z|^2
im not sure what the star means. should i ignore this?

any help on the differentiation would be great im not sure what to do with the negative fraction powers

differentiate w = 6t^(-3/2) - 4t^(3/2)
You do it just like any other derivative:
$\displaystyle w^{\prime} = 6 \cdot \left ( -\frac{3}{2} \right )t^{-5/2} - 4 \cdot \left ( \frac{3}{2} \right ) t^{1/2}$

-Dan

and given that z = 1 + j and w = 2-3j find

w/z
yes its a complex number division the equation i have is wz*/|z|^2
im not sure what the star means. should i ignore this?
Owwwwch!

The star means to take the complex conjugate of the variable. So since $\displaystyle z = 1 + j$, $\displaystyle z^* = 1 - j$.

Here it is in its full glory:
$\displaystyle \frac{w}{z} = \frac{2 - 3j}{1 + j}$

The problem with leaving the expression like this is that there is a radical in the denominator, $\displaystyle j = \sqrt{-1}$, which is typically removed. So if you multiply the denominator by the conjugate of the denominator (in this case rechristened as the "complex conjugate") we get:
$\displaystyle \frac{w}{z} = \frac{2 - 3j}{1 + j}$

$\displaystyle = \frac{2 - 3j}{1 + j} \cdot \frac{1 - j}{1 - j} = \frac{(2 - 3j)(1 - j)}{1^2 - j^2}$

$\displaystyle = \frac{2 - 2j - 3j + 3j^2}{1 - (-1)} = \frac{2 - 5j + 3(-1)}{1 + 1}$

$\displaystyle = \frac{-1 - 5j}{2}$

-Dan

7. Star? What Star?
The conjugate of z is $\displaystyle {\overline z }$.
Does that appear as z* in your browser? If so, what is the browser?

8. ## star

Originally Posted by Plato
Star? What Star?
The conjugate of z is $\displaystyle {\overline z }$.
Does that appear as z* in your browser? If so, what is the browser?
No i am using explorer but on the formulae sheet it is an star not an overline. I think they mean the same thing.

No i am using explorer but on the formulae sheet it is an star not an overline. I think they mean the same thing.
I do know that Europeans often use j in the same way that north American use i. (I did study a year at the University of Manchester UK) but I did not know about using * for conjugate. So yes you are correct about that.

10. Originally Posted by Plato
I do know that Europeans often use j in the same way that north American use i. (I did study a year at the University of Manchester UK) but I did not know about using * for conjugate. So yes you are correct about that.
The asterisk (what you are calling a star) is commonly used in wave mechanics in Physics.

-Dan

11. Originally Posted by Plato
I do know that Europeans often use j in the same way that north American use i. (I did study a year at the University of Manchester UK) but I did not know about using * for conjugate. So yes you are correct about that.
1. European mathematicians use i exclusively for the imaginary unit. Engineers
of the electrical/electronic persuasion use j for the imaginary unit to avoid
confusion with the use of i for current (I had to go through the last paper I
sent for publication changing all the i's to j's as it was going to a
nominally electronic engineering journal, though as most of the stuff
published in it is related to DSP I think they are living in the past).

2. Often * is used to denote complex conjugate plain in ASCII maths.

RonL