please do it perfectly

Question:.
Solution: Let z = x+ iy is any complex number and is written in polar form as
z = r.cos
q
+ i r.sin
q
(where r is modulus and
q
is amplitudes of the complex number z)
now
1
z
= (r.cos
q
+ i r.sin
q
)-1
= r-1 (cos
q
+ i.sin
q
)-1
=
1
r
(cos
q
- i.sin
q
) (using de moivre’s theorem)
According to de moivre’s theorem
(cos
q
+ i.sin
q
)n = cos(n
q
) + i.sin(n
q
)
z +
1
z
= r.cos
q
+ i r.sin
q
+
1
r
(cos
q
- i.sin
q
)
=(r +
1
r
) cos
q
+ i (r -
1
r
) sin
q
Answer: f(z ) = z +
1
z
= (r +
1
r
) cos
q
+ i (r -
1
r
) sin
q
    2.(a) f(z) = iz + 2
f’(z) = i , which exist everywhere
Answer: f”(z) = 0
(b) f(z) = e-xe-iy
=e(x+iy)
= e-z
f’(z) = - e-z , which exist everywhere
f”(z) = e-z
(c) f(z) = z3
f’(z) = 3 z2 , which exist everywhere
Answer: f”(z) = 3 z2
(d) f(z) = cosx coshy – isinx sinhy
= cos(x+iy)
= cosz
f’(z) = -sinz , which exist everywhere
f”(z) = -cosz = -
Answer: f”(z) = - f(z)
    To find f’(z)
(a) f(z) =
4
1
z
Answer: f’(z) =
5
4
z
-
, which exist everywhere for all z
¹
0
(b) f(z) =
/2
i
re
q
=
(
)
1/2
i
re
q
= z1/2
f’(z) =
1
2
z
Answer: f’(z) =
1
2()
fz
(c) f(z) =
(
)
cos(log)sin(log)
erir
q
-
+
=
log
ir
ee
q
-
=
log
ir
e
q
-+
=
(
)
log
iir
e
q
+
We know that z =
i
re
q

Þ
ln(z) = ln(r) + i
q
=
(
)
log
iz
e

f’(z) =
(
)
log
.
iz
i
e
z
Answer: f’(z) =
().
i
fz
z
    To check wether the function is analytic or not
(a) f(z) = xy + iy = u+ iv
u = xy and v = y
0
1
uv
yand
xx
uv
xand
yy
¶¶
==
¶¶
¶¶
==
¶¶
Thus we see that cauchy-Riemann conditions
u
x
¶
¶
=...