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Evaluating complex expression

shen07

Member
Aug 14, 2013
54
How to go about solving this one?

(1-i)-1+i

so far i have tries using the famous Uv=ev(ln(U)

im stuck where i get

e-1/2ln(2)+(pi/4)-2kpi*ei(-1/2ln(2)+(pi/4)-2kpi)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle z=(1-i)^{i-1}\)

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

Hence:

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

shen07

Member
Aug 14, 2013
54
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:

\(\displaystyle z=(1-i)^{i-1}\)

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

Hence:

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



Yeah thats right but however i have to include the 2K\(\displaystyle {\pi}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yeah thats right but however i have to include the 2K\(\displaystyle {\pi}\)
I see this as wholly unnecessary, but write it in if it is required.
 

shen07

Member
Aug 14, 2013
54
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..
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle z=(1-i)^{i-1}\)

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

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle z=(1-i)^{i-1}\)

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

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

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

Ofcourse we could do

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

But this will be not complete since it only gives the solution when \(\displaystyle k=0\).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
So then the complete solution is:

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

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

Member
Aug 14, 2013
54
Also i would like to add something

e2k\(\displaystyle {\pi}\)i=cos(2k\(\displaystyle {\pi}\))+isin(2k\(\displaystyle {\pi}\))= 1

\(\displaystyle {\forall} k {\in} {\mathbb{z}}\)
but

e2k\(\displaystyle {\pi}\)\(\displaystyle {\neq}\)1