78
5 The many-valued functions a
zand z
aExample 5.1 Compute all the complex values ofzi forz∈C\ {0}.
We get by insertion into the formula of the many-valued function za fora=ithat zi := exp(i{ln|z|+i(Arg z−2pπ)})
= e−Argz+2pπcot{cos(ln|z|) +i·sin(ln|z|)}, p∈Z.
We have here used that we also have arg ={Argz−2pπ|p∈Z}which give a nicer final result.
Example 5.2 Find all values of (a) ii, (b) 1i, (c) 1√2.
Here we shall stick very close to the definitions.
(a)
ii:= exp(ilogi) = exp
i
ln|i|+i π
2 −2pπ
= exp −π
2 + 2pπ
, p∈Z, which is a sequence of real numbers!
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(b)
1i:= exp(ilog 1) = exp(i· {0−2ipπ}) =e2pπ, p∈Z, which is also a sequence of real numbers.
(c)
1√2:= exp(√
2·log 1) = exp(√
2·2pπi) = cos(2√
2pπ) +isin(2√
2pπ), p∈Z. This set is dense on the unit circle.
Remark 5.1 The examples above show that concerning many-valued functions one cannot rely on one’s intuition. ♦
Example 5.3 Givenc∈C\ {0} andz=x+iy∈C. (a) Find all values of log (cz).
(b) Find all values ofz logc.
(c) For which values ofz andc do we havelog (cz) =zlogc?
(a) Letp,q∈Zdenote free parameters. Then log (cz) = log ez·logc
=z·logc+ 2ipπ= (x+iy){ln|c|+iArgc+ 2iqπ}+ 2ipπ
= xln|c|−yArgc−2yqπ+i{xArgc+y ln|c|+2xqπ+2pπ}.
(b) It follows from the computation above thatz·logcis obtained by puttingp= 0, thus z logc=xln|c|−yArgc−2yqπ+i{zArgc+yln|c|+2xqπ}, q∈Z.
(c) A necessary (though not sufficient) condition for log (cz) =zlogc,
is that the sets
{2π(xq+p)|p, q∈Z} and {2πxq|q∈Z}
are identical.
Hence a necessary and sufficient condition is that Z{xq|q∈Z}, thusx∈Q.
If y = 0, then the real parts are identical, if and only if we choose the same q ∈ Z. Then, concerning the imaginary parts, we are forced to choose p = 0, and p ∈ Z is no longer a free parameter. Therefore,y= 0.
When this is the case, i.e.z=x∈Q, then
log (xx) =xln|c|+i{xArgc+ (xq+p)·2π}, p, q∈Z,
80 and
xlogc=xln|c|+i{xArgc+xq·2π}, p∈Z.
Since {xq+p | p, q ∈ Z} considered as a set (not counted by multiplicity) is identical with {xq|q∈Z}, the two expressions are identical, if and only if
z=x∈Q,
whilec∈C\ {0}can be chosen arbitrarily.
Example 5.4 Compute all values of
(a) log(1 +i)πi, (b) (−i)−i, (c) 3π, (d) 2πi.
(a) We first compute (1 +i)πi. Here,
(1 +i)πi = exp(πilog(1 +i)) = exp
πi 1
2 ln 2 +i π
4 −2pπ
= exp
−π2
4 + 2pπ2+iπ 2 ln 2
, p∈Z, hence
log(1 +i)πi=
2p−1 4
π2+iπ 1
2 ln 2 + 2q
, p, q∈Z.
Remark 5.2 We see that we havetwoindependent parametersp, q∈Z. ♦ (b) A computation gives here
(−i)−i = exp(−i{ln| −i|+iarg(−i)}) = exp(−i·i·arg(−i))
= exp(arg(−i)) = exp
2p−1 2
π
, p∈Z, i.e. a set of real positive numbers.
(c) Here,
3π= exp(πln 3) = exp(π{ln 3 + 2ipπ}) =eπ3
cos 2pπ2
+isin 2pπ2
, p∈Z. It is easily seen that this set is dense on the circle of centrum 0 and radiuseπln 3. (d) Here,
2πi= exp(πilog 2) = exp(πi{ln 2−2ipπ}) =e2pπ2{cos(πln 2) +i sin(πln 2)}, p∈Z, which represent infinitely many numbers on the half line from 0 and forming the angleπln 2 with the real axis.
0 0.2 0.4 0.6 0.8
–0.4 –0.2 0.2 0.4
Figure 36: The points of(d)lie on a half line.
Example 5.5 Compute all values of
(a) (1 +i)1+i, (b) (1 +i)i(1 +i)−i, (c) i2.
(a) It follows from the definition and a computation that (1 +i)1+i := exp((1 +i) log(1 +i)) = exp
(1 +i)
1
2 ln 2 +i π
4 + 2pπ
= exp 1
2 ln 2−π
4 −2pπ+i 1
2 ln 2 + π 4 + 2pπ
= √
2 exp −π
4 −2pπ ×
cos
1
2 ln 2 +π 2
+isin
1
2 ln 2 +π 2
.
(b) By the definition,
(1 +i)i(1 +i)−i := exp(ilog(1 +i))·exp(−ilog(1 +i))
= exp
i 1
2 ln 2 +i π
4 + 2pπ
exp
−i 1
2 ln 2 +i π
4 + 2qπ
= exp
−π
4 −2pπ+i1
2 ln 2−i1
2 ln 2 +π 4 + 2qπ
= exp(2(q−p)π) = exp(2nπ), n∈Z.
(c) At this stage the reader should be suspicious concerning many-valued functions. Nevertheless, this example should not give any problem,
i2=−1.
82 Example 5.6 Compute all values of
(a) 2i, (b) √4
16, (c) (√
3−i)i, (d) (−1)i.
(a)
2i:= exp(ilog 2) = exp(i(ln 2−2ipπ)) =e2pπ{cos(ln 2) +isin(ln 2)}, p∈Z. (b)
√4
16 := exp 1
4 log 16
= exp 1
4(4 ln 2 + 2ipπ)
= exp
ln 2 +i pπ 2
={2, 2i,−2,−2i}.
(c) (√
3−i)i ; = exp(i log(√
3−i)) = exp
i
ln 2 +i −π
6 −2ipπ
= exp π
6 + 2pπ
· {cos(ln 2) +isin(ln 2)}, p∈Z.
(d)
(−1)i=|exp(i{−iπ−2ipπ})|=|exp(π+ 2pπ)|=e(2p+1)π, p∈Z.
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Remark 5.3 The strange thing here is that also
(−1)i= exp(i{−iπ−2ipπ}) = exp(π+ 2pπ) =e(2p+1)π, p∈Z, hence (−1)i=(−1)i. ♦
Example 5.7 Putz=r eiθ. Compute all values of (a) Re zi
, (b) Im zi
, (c) zi. It follows from
zi := exp(i{lnr+iθ−2ipπ}) = exp(−θ+ 2pπ+ilnr), that
(a)
Re zi
=eθ+2pπcos(lnr), p∈Z, (b)
Im zi
=e−θ+2pπsin(lnr), p∈Z, (c) zi=e−θ+2pπ, p∈Z.
Example 5.8 Putz=x+iy. Compute all values of (a) Re(iz), (b) Im(iz), (c) |iz|. It follows from
iz := exp(z·logi) = exp
(x+iy)i π
2 + 2pπ
= exp
−u 1
2 + 2p
π+i x 1
2 + 2p
π
, p∈Z, that
(a)
Re (iz) = exp
−yπ
2p+1 2
·cos
xπ
2p+1 2
, p∈Z, (b)
Im (iz) = exp
−yπ
2p+1 2
·sin
xπ
2p+1 2
, p∈Z,
84 (c)
|iz|= exp
−yπ
2p+1 2
, p∈Z.
Example 5.9 Find all the solutions of the following equations (a) Logz= 1
4πi, (b) ez=i, (c) ez= 1 +i√ 3.
(a) We first note that 1
4πi lies in the image of the principal logarithm, so there exists a solution.
This is given by z=eLogz= exp
iπ
4
= 1
√2+i 1
√2 = 1
√2(1 +i).
(b)
z= logi=i π
2 + 2pπ
, p∈Z. (c)
z= log(1 +i√
3) = ln 2 +i π
3 + 2pπ
, p∈Z.
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© Deloitte & Touche LLP and affiliated entities.