Complex Analysis - Mapping of a straight line and Principal value

**fourierT** · December 11th 2013, 06:09 PM

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.

**romsek** · December 11th 2013, 06:52 PM

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 $\small 2^{i+1} \text { for } (0 \leq \theta <2\pi)$

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

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

**fourierT** · December 11th 2013, 07:07 PM

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

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

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

?

**romsek** · December 11th 2013, 07:19 PM

Originally Posted by fourierT

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

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

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

?

no...

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

you could have started with

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

I guess that's a bit quicker

**fourierT** · December 11th 2013, 07:27 PM

Originally Posted by romsek

no...

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

you could have started with

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

I guess that's a bit quicker

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

**romsek** · December 11th 2013, 07:34 PM

Originally Posted by fourierT

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

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

**fourierT** · December 11th 2013, 07:44 PM

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

**romsek** · December 11th 2013, 07:57 PM

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

that's all it is.

Math Help - Complex Analysis - Mapping of a straight line and Principal value

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