MHB Solving Complex Expressions for Scientists

[shen07]

How to go about solving this one?

(1-i)^-1+i

so far i have tries using the famous U^v=e^v(ln(U)

im stuck where i get

e^{-1/2ln(2)+(pi/4)-2kpi}*e^{i(-1/2ln(2)+(pi/4)-2kpi)}

 

[MarkFL]

You have a good approach, but I think you have made an algebraic slip somewhere. To avoid a complex argument for the natural log function, I wrote:

$$z=(1-i)^{i-1}$$

$$1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4} \right)+i\sin\left(-\frac{\pi}{4} \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}$$

Hence:

$$z=\left(e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i} \right)^{i-1}=e^{\frac{\pi}{4}-\frac{1}{2}\ln(2)}e^{\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)i}=\frac{e^{\frac{\pi}{4}}}{\sqrt{2}}\left(\cos\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right) \right)$$

 

[shen07]

You have a good approach, but I think you have made an algebraic slip somewhere. To avoid a complex argument for the natural log function, I wrote:

$$z=(1-i)^{i-1}$$

$$1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4} \right)+i\sin\left(-\frac{\pi}{4} \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}$$

Hence:

$$z=\left(e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i} \right)^{i-1}=e^{\frac{\pi}{4}-\frac{1}{2}\ln(2)}e^{\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)i}=\frac{e^{\frac{\pi}{4}}}{\sqrt{2}}\left(\cos\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right) \right)$$

Yeah that's right but however i have to include the 2K$${\pi}$$

 

[MarkFL]

Yeah that's right but however i have to include the 2K$${\pi}$$

I see this as wholly unnecessary, but write it in if it is required.

 

[shen07]

I see this as wholly unnecessary, but write it in if it is required.

You can find more values if you take one turn or several turns..My Professor told m that..

 

[alyafey22]

I see this as wholly unnecessary, but write it in if it is required.

I think it becomes necessary if we study complex numbers , Of course we always consider the Principle Logarithm if we are working on the real numbers .

 

[MarkFL]

$$z=(1-i)^{i-1}$$

$$1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}+2k\pi \right)+i\sin\left(-\frac{\pi}{4}+2k\pi \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i+2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}e^{2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}$$

This is why I found it unnecessary in this case. Am I missing something?

 

[alyafey22]

$$z=(1-i)^{i-1}$$

$$1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}+2k\pi \right)+i\sin\left(-\frac{\pi}{4}+2k\pi \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i+2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}e^{2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}$$

This is why I found it unnecessary in this case. Am I missing something?

I think the OP talks about the logarithm since $a^b$ is actually a multivalued function in the sense that b is a complex number .

Consider the simpler example

$$\Large i^i = e^{i\log(i)}=e^{i(\ln(1)+(\frac{\pi}{2}+2k\pi) i )}=e^{-\frac{\pi}{2}-2k\pi }$$

Ofcourse we could do

$$ \Large i^i= (e^{i\frac{\pi}{2}})^i=e^{-\frac{\pi}{2}}$$

But this will be not complete since it only gives the solution when $$k=0$$.

 

[MarkFL]

So then the complete solution is:

$$z=\frac{e^{\frac{\pi}{4}(8k+1)}}{\sqrt{2}}\left( \cos\left(\frac{\pi}{4}(8k+1)+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}(8k+1)+\frac{1}{2}\ln(2) \right) \right)$$ ?

W|A told me only when $k=0$ is this a solution. (Giggle)

My apologies for any confusion I caused to the OP.(Doh)

 

[shen07]

Also i would like to add something

e^2k$${\pi}$$i=cos(2k$${\pi}$$)+isin(2k$${\pi}$$)= 1

$${\forall} k {\in} {\mathbb{z}}$$
but

e^2k$${\pi}$$$${\neq}$$1