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Complex Analysis Logarithm Identities

1. The document contains proofs of 7 identities involving logarithms of complex numbers. It proves that: log(z1z2) = logz1 + logz2; arg(z1z2) = argz1 + argz2; and ln|z1z2| + iarg(z1z2) = (ln|z1| + iargz1) + (ln|z2| + iargz2). 2. It also proves that log(z1/z2) = logz1 - logz2 and that if z is nonzero, then z^n = e^nlogz for any value of logz.

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88 views4 pages

Complex Analysis Logarithm Identities

1. The document contains proofs of 7 identities involving logarithms of complex numbers. It proves that: log(z1z2) = logz1 + logz2; arg(z1z2) = argz1 + argz2; and ln|z1z2| + iarg(z1z2) = (ln|z1| + iargz1) + (ln|z2| + iargz2). 2. It also proves that log(z1/z2) = logz1 - logz2 and that if z is nonzero, then z^n = e^nlogz for any value of logz.

Uploaded by

raddy.tahil
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Republic of the Philippines

Western Mindanao State University


College of Science and Mathematics
DEPARTMENT OF MATHEMATICS AND STATISTICS

MATH 153
COMPLEX ANALYSIS
Activity 5
Date: 10/11/2022

Jaafar, Fatima Shelaine A.


Buenavista, Jerico M.
BS – Mathematics IV

Prove the following identities


Some identities involving Logarithms
If 𝑧1 and 𝑧2 denote any two nonzero complex numbers,

1. 𝑙𝑜𝑔(𝑧1 𝑧2 ) = 𝑙𝑜𝑔𝑧1 + 𝑙𝑜𝑔𝑧2


Proof.

Let 𝑧1 = 𝑟1 𝑒 𝑖𝜃1 and 𝑧2 = 𝑟2 𝑒 𝑖𝜃2


Thus,
𝑧1 𝑧2 = (𝑟1 𝑒 𝑖𝜃1 )(𝑟2 𝑒 𝑖𝜃2 )
= (𝑟1 𝑟2 )𝑒 𝑖(𝜃1+𝜃2 )
Therefore,
𝑙𝑜𝑔(𝑧1 𝑧2 ) = 𝑙𝑛(𝑟1 𝑟2 ) + 𝑖((𝜃1 + 𝜃2 ) + 2𝜋𝑛) (𝑛 =, ±1, ±2, ±3, … )
= 𝑙𝑛(𝑟1 𝑟2 ) + 𝑖((𝜃1 + 𝜃2 ) + 4𝜋𝑛)
= 𝑙𝑛(𝑟1 𝑟2 ) + 𝑖((𝜃1 + 𝜃2 ) + (2𝜋𝑛 + 2𝜋𝑛))
= 𝑙𝑛𝑟1 + 𝑙𝑛𝑟2 + 𝑖 (𝜃1 + 2𝜋𝑛) + (𝜃2 + 2𝜋𝑛)
= (𝑙𝑛𝑟1 + 𝑖 (𝜃1 + 2𝜋𝑛)) + (𝑙𝑛𝑟2 + 𝑖 (𝜃2 + 2𝜋𝑛))
𝑙𝑜𝑔(𝑧1 𝑧2 ) = 𝑙𝑜𝑔𝑧1 + 𝑙𝑜𝑔𝑧2 . ∎

2. 𝑎𝑟𝑔(𝑧1 𝑧2 ) = 𝑎𝑟𝑔𝑧1 + 𝑎𝑟𝑔𝑧2


Proof.

Let 𝑧1 = 𝑟1 𝑒 𝑖𝜃1 and 𝑧2 = 𝑟2 𝑒 𝑖𝜃2


Thus,
𝐴𝑟𝑔𝑧1 = 𝜃1 , 𝐴𝑟𝑔𝑧2 = 𝜃2
𝐴𝑟𝑔(𝑧1 𝑧2 ) = (𝜃1 + 𝜃2 )
Then,
𝑎𝑟𝑔𝑧1 = 𝐴𝑟𝑔𝑧1 + 2𝜋𝑛 = 𝜃1 + 2𝜋𝑛 (𝑛 =, ±1, ±2, ±3, … )
𝑎𝑟𝑔𝑧2 = 𝐴𝑟𝑔𝑧2 + 2𝜋𝑛 = 𝜃2 + 2𝜋𝑛 (𝑛 =, ±1, ±2, ±3, … )
Now,
𝑧1 𝑧2 = (𝑟1 𝑒 𝑖𝜃1 )(𝑟2 𝑒 𝑖𝜃2 ) == (𝑟1 𝑟2 )𝑒 𝑖(𝜃1+𝜃2 )
Since 𝐴𝑟𝑔(𝑧1 𝑧2 ) = (𝜃1 + 𝜃2 )
Therefore,
𝑎𝑟𝑔(𝑧1 𝑧2 ) = 𝐴𝑟𝑔(𝑧1 𝑧2 ) + 2𝜋𝑛
= (𝜃1 + 𝜃2 ) + 2𝜋𝑛
= (𝜃1 + 𝜃2 ) + 4𝜋𝑛
= (𝜃1 + 𝜃2 ) + (2𝜋𝑛 + 2𝜋𝑛)
= (𝜃1 + 2𝜋𝑛) + (𝜃2 + 2𝜋𝑛)
𝑎𝑟𝑔(𝑧1 𝑧2 ) = 𝑎𝑟𝑔𝑧1 + 𝑎𝑟𝑔𝑧2 . ∎

3. 𝑙𝑛|𝑧1 𝑧2 | + 𝑖𝑎𝑟𝑔(𝑧1 𝑧2 ) = (𝑙𝑛|𝑧1 | + 𝑖𝑎𝑟𝑔𝑧1 ) + (𝑙𝑛|𝑧2 | + 𝑖𝑎𝑟𝑔𝑧2 )


Proof.

Since 𝑙𝑛|𝑧1 𝑧2 | = 𝑙𝑛|𝑧1 | + 𝑙𝑛|𝑧2 | and by problem #2 𝑎𝑟𝑔(𝑧1 𝑧2 ) = 𝑎𝑟𝑔𝑧1 + 𝑎𝑟𝑔𝑧2 , we get
𝑙𝑛|𝑧1 𝑧2 | +i 𝑎𝑟𝑔(𝑧1 𝑧2 ) = (𝑙𝑛|𝑧1 | + 𝑙𝑛|𝑧2 |) + 𝑖(𝑎𝑟𝑔𝑧1 + 𝑎𝑟𝑔𝑧2 )
𝑙𝑛|𝑧1 𝑧2 | +i 𝑎𝑟𝑔(𝑧1 𝑧2 ) = (𝑙𝑛|𝑧1 | + 𝑖𝑎𝑟𝑔𝑧1 ) + (𝑙𝑛|𝑧2 | + 𝑖𝑎𝑟𝑔𝑧2 ). ∎

𝑧1
4. 𝑙𝑜𝑔 = 𝑙𝑜𝑔𝑧1 − 𝑙𝑜𝑔𝑧2
𝑧2
Proof.

First to show that,


1
𝑙𝑜𝑔 = − log(𝑧2 )
𝑧2

Then, Let 𝑧2 = 𝑟2 𝑒 𝑖𝜃2 , (𝑛 =, ±1, ±2, ±3, … )


1 1
𝑙𝑜𝑔 = 𝑙𝑜𝑔
𝑧2 𝑟2 𝑒 𝑖𝜃2
= log(𝑟2 −1 𝑒 −𝜃2 𝑖 )
= ln(𝑟2 −1 ) + 𝑖(-𝜃2+ 2𝜋𝑛)
= − ln(𝑟2 ) − 𝑖(𝜃2 − 2𝜋𝑛)
= −(ln(𝑟2 ) + 𝑖(𝜃2 − 2𝜋𝑛))
1
𝑙𝑜𝑔 = − log(𝑧2 ).
𝑧2
Now, Let 𝑧1 = 𝑟1 𝑒 𝑖𝜃1 and 𝑧2 = 𝑟2 𝑒 𝑖𝜃2 , (𝑛 =, ±1, ±2, ±3, … )
Hence,
1
𝑙𝑜𝑔 = − log(𝑧2 )
𝑧2
Thus, by using relation, we get
𝑧1 1
𝑙𝑜𝑔 = log (𝑧1 ( ))
𝑧2 𝑧2
1
= log(𝑧1 ) + log (𝑧 )
2
𝑧1
𝑙𝑜𝑔 = 𝑙𝑜𝑔𝑧1 − 𝑙𝑜𝑔𝑧2 . ∎
𝑧2

5. If 𝑧 is a nonzero complex number, then 𝑧 𝑛 = 𝑒 𝑛𝑙𝑜𝑔𝑧 (𝑛 =, ±1, ±2, ±3, … ) for any value of 𝑙𝑜𝑔𝑧 that is
taken.
Proof.

Let 𝑧 be a nonzero complex number, thus 𝑧 = 𝑒 𝑙𝑜𝑔𝑧


Now,
𝑛
𝑧 𝑛 = (𝑒 𝑙𝑜𝑔𝑧 ) = 𝑒 𝑛𝑙𝑜𝑔𝑧 , (𝑛 =, ±1, ±2, ±3, … ). ∎

6. It is true that when 𝑧 ≠ 0,


1 1
𝑧 𝑛 = 𝑒 𝑛𝑙𝑜𝑔𝑧 , (𝑛 = 1,2,3, … )
Proof.

Let 𝑧 be a nonzero complex number, thus 𝑧 = 𝑟𝑒 𝑖𝜃


Then,
1 1
𝑧 𝑛 = (𝑟𝑒 𝑖𝜃 )𝑛
1 𝑖𝜃
= 𝑟𝑛𝑒 𝑛
1 𝑖(𝜃+2𝜋𝑘)
= 𝑟𝑛𝑒 𝑛
1 𝑖(𝜃+2𝜋𝑘)
= 𝑒 𝑙𝑛𝑟 𝑛 𝑒 𝑛
1 𝑖(𝜃+2𝜋𝑘)
+
= 𝑒 𝑙𝑛𝑟 𝑛 𝑛

1 𝑖(𝜃+2𝜋𝑘)
= 𝑒 𝑛𝑙𝑛𝑟+ 𝑛
1
= 𝑒 𝑛(𝑙𝑛𝑟+𝑖(𝜃+2𝜋𝑘)) , (𝑘 =, ±1, ±2, ±3, … )
1 1
𝑧 𝑛 = 𝑒 𝑛𝑙𝑜𝑔𝑧 , (𝑛 = 1,2,3, … ). ∎

1 𝜃 2𝜋𝑘
7. 𝑒 𝑛𝑙𝑜𝑔𝑧 = 𝑛√𝑟𝑒 𝑖(𝑛+ 𝑛
)
, (𝑘 =, ±1, ±2, ±3, … )
Proof.

1 1
(𝑙𝑛𝑟+𝑖(𝜃+2𝜋𝑘))
𝑒 𝑛𝑙𝑜𝑔𝑧 = 𝑒 𝑛
1 𝑖(𝜃+2𝜋𝑘)
= 𝑒 𝑛𝑙𝑛𝑟+ 𝑛
1 𝑖(𝜃+2𝜋𝑘)
= 𝑒 𝑙𝑛𝑟 𝑛 𝑒 𝑛
1 𝑖(𝜃+2𝜋𝑘)
= 𝑟 𝑛𝑒 𝑛
1 𝜃 2𝜋𝑘
𝑖( + )
𝑒 𝑛𝑙𝑜𝑔𝑧 = √𝑟𝑒
𝑛
𝑛 𝑛 , (𝑘 =, ±1, ±2, ±3, … ). ∎

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