Thread: Complex Analysis - Mapping of a straight line and Principal value

I've done part a) I just simply cubed z and expanded, then seperated real and imaginary. I assume this is all that's needed?

For part b) I have no idea how to do it.

Originally Posted by fourierT

I've done part a) I just simply cubed z and expanded, then seperated real and imaginary. I assume this is all that's needed?

For part b) I have no idea how to do it.

I doubt they are talking about the Cauchy Principal value so
what I'm going to assume they mean in (b) is simply the value of $\displaystyle \small 2^{i+1} \text { for } (0 \leq \theta <2\pi)$

$\displaystyle \small 2^{i+1}=e^{(i+1)ln(2)}$

$\displaystyle \text{You can probably come up with } z=r e^{i\theta}\text{ for that.}$

$\displaystyle \small 2^{i+1}=e^{(i+1)ln(2)}$

$\displaystyle \text{You can probably come up with } z=r e^{i\theta}\text{ for that.}$

Is it cos(ln2) + (i+1)sin(ln2)

?

Originally Posted by fourierT

$\displaystyle \small 2^{i+1}=e^{(i+1)ln(2)}$

$\displaystyle \text{You can probably come up with } z=r e^{i\theta}\text{ for that.}$

Is it cos(ln2) + (i+1)sin(ln2)

?

no...

$\displaystyle e^{(i+1)ln(2)} = e^{ln(2) + ln(2) i} = 2 * e^{ln(2) i} = 2(cos(ln(2))+i sin(ln(2)))$

you could have started with

$\displaystyle 2^{i+1}=2*2^{i}=2e^{ln(2)i}$

I guess that's a bit quicker

Originally Posted by romsek

no...

$\displaystyle e^{(i+1)ln(2)} = e^{ln(2) + ln(2) i} = 2 * e^{ln(2) i} = 2(cos(ln(2))+i sin(ln(2)))$

you could have started with

$\displaystyle 2^{i+1}=2*2^{i}=2e^{ln(2)i}$

I guess that's a bit quicker

How can we evaluate that in algebraic form..wouldn't it just be numbers.

Originally Posted by fourierT

How can we evaluate that in algebraic form..wouldn't it just be numbers.

you see any variables in $\displaystyle 2^{i+1}$ ?

Sure, i is algebraic, but is that all it wants, the real part and the imaginary part?

the algebraic form of a complex number is z = x + i y

that's all it is.