SC105- Calculus and Complex Variables
Home Work 7
Week 8: October 12, 2015
Tutorial Discussion Week: October 12, 2015
Tutors: Krishna Gopal Benerjee and Dixita Limbachiya
(1) Find the complex solution to each of the following equations:
(a). ew = i
S::
For z = i , |z| = 1 and Arg(z) = π2 + 2nπ
⇒ w = log(i) = loge (1) + i( π2 + 2nπ)
i , n ∈ Z.
⇒ w = (4n+1)π
2
7π
⇒ w = . . . , − 2 i, − 3π
i, π2 i, 5π
i, . . .
2
2
w
⇒ for these w, e = i
(b). ew = i + 1
S::
√
For z = 1 + i , |z| = 2 and Arg(z) = π4 + 2nπ
√
⇒ w = log(1 + i) = loge ( 2) + i( π4 + 2nπ)
√
but loge 2 = 12 loge 2 ⇒ w = 12 loge 2 + (8n+1)π
i,
4
n ∈ Z.
(c). ew = −2
s::
For z = −2 , |z| = 2 and Arg(z) = π + 2nπ
⇒ w = loge −2 = loge 2 + i(π + 2nπ) n ∈ Z.
(2) Compute the principal value of the complex logarithm log(z) for:
(a). z = i
S::
For z = i , |z| = 1 and Arg(z) =
⇒ log(i) = loge 1 + π2 = π2 i.
π
2
1
(b). z = i + 1
S::
√
For z = 1 + i , |z| = 2 and Arg(z) = π4
and so,
√
log(1 + i) =
loge 2 + π4 i
1
loge 2 + π4 i
=
2
≈ 0.3466 + 0.7854i
(c). z = −2
S::
For z = −2 , |z| = 2 and Arg(z) = π
⇒ log(−2) = loge 2 + πi ≈ 0.6931 + 3.1416i
(3) Find the image of the annulus 2 ≤ |z| ≤ 4 under the logarithmic mapping w = log(z).
S::
The boundry circles |z| = 2 & |z| = 4 maps onto the vertical line segments,
u = loge 2 & v = loge 4 , −π < v ≤ π
and each circle |z| = r, 2 ≤ r ≤ 4 maps on to the vertical line segment,
v = loge r, −π < r ≤ π.
Since the real logarithmic function is increasing on its domain,
⇒ v = loge r takes on all values in the inteval,
loge 2 ≤ v ≤ loge 4
when 2 ≤ r ≤ 4.
⇒ The image of the annulus 2 ≤ |z| ≤ 4 is the rectangular region,
loge 2 ≤ 4 ≤ loge 4, −π < v ≤ π
See the figure-1 given below,
2
(4) Find the values :
(a). (i)2i s::
Note since,
log(i) =
(4n + 1)π
i
2
∴ if z = i and c = 2i
πi
(i)2i = e2ilog(i) = e2i[(4n+1) 2 ] = e−(4n+1)π ,
where n ∈ Z
(b). (i + 1)i
s::
1
(8n + 1)π
∵ log(1 + i) = loge 2 +
i
2
4
⇒ (1 + i)i =
exp(i loge (1 + i))
= exp(i[loge 2/2 + (8n + 1)πi/4])
i
⇒ (1 + i) =
exp(− (8n+1)π
+ i log2e 2 )
n∈Z
4
(5) Find the principal value of each complex power:
3
i
(a). (−3) π
S::
For z = −3 , |z| = 3 , & Arg(−3) = π
and so log(−3) = loge 3 + iπ
⇒ (−3)i/π =
exp( πi · loge (−3))
=
exp( πi (loge 3 + iπ))
=
exp(−1 + i(loge 3)/π)
−1
cos( logπe 3 ) + i sin( logπe 3 )
= e
⇒ (−3)i/π ≈
0.3456 + 0.1260i
(b). (2i)1−i
For z = 2i, |z| = 2Arg(z) =
π
2
iπ
2
⇒ log(2i) =
loge 2 +
⇒ (2i)1−i =
e(1−i) log(2i)
π
e(1−i)(loge 2+i 2 )
=
π
π
eloge 2+ 2 −i(loge 2− 2 )
⇒ (2i)1−i =
π
⇒ (2i)1−i = eloge 2+ 2 [cos(logc 2 − π2 ) − i sin(loge 2 − π2 )]
∼
=
6.1474 + 7.400i
4
√
(6) Find all values of sin−1 ( 5).
∵ sin−1 (z)
⇒ sin−1
√
5
=
√
√
1
= −i loge [i 5 + (1 − ( 5)2 ) 2 ]
=
√
1
−i loge [i 5 + (−4) 2
=
√
−i loge [i 5 ± 2i]
=
√
−i loge [( 5 ± 2)i]
√
but[( 5 ± 2)i]
=
√
arg[( 5 ± 2)i]
=
√
⇒ log[( 5 ± 2)i] =
⇒ sin−1
√
5
√
∵ loge ( 5 ± 2)
1
−i loge [iz + (1 − z 2 ) 2 ]
=
=
√
5±2
π
2
loge
√
5 ± 2 + i( π2 + 2nπ)
(4n+1)π
2
√
± i loge ( 5 + 2)
√
± loge ( 5 + 2)
5
n∈Z