1. a) Z1 = 2-5i, z2 = 1 =2i
Z1 = 22 + 52 = 4 +25 = 29
b) = (2-5i)(1+2i) = 2 + 4i-5i +10 = 12-i
c) (2 = {(2-5i) + [3(1 +2i)]}2
= {(2-5i + 3 +6i}2
= {(5 +i}2
=25 +10i – 1
=24 + 10i
d) {2 = +2(1+) + (1 + )2
= (2-5i)2 = 4- 10i +25 = 29-10i
2(1+) = 2(2-5i)(1 +1 +2i) =(4- 10i)(2 +2i) = 3 + 8i -2oi + 20
= 28- 12i
(1 + )2 = (1 + 1 =2i)2 =(2 +2i)2 = 4 + 8i- 4 = 8i
{2 = 29-10i + 28- 12i + 8i = 57-14i
2. a) = =
= +
b) (2 +i)3 = 23 + 3(22i) + 3(2i2) +i3 = 8 +12i-6-I = 2 + 11i
c) 3 + = 3i + i
= + i
=15 +6i
=21 i
d) - = = =
3. a)
z = +i and w =1 +
Z =r[cosθ + isinθ]
θ = =
r = =2
z = 2[cos +isin]
w = r1 (cos α + isinα)
r = =2
θ = =
w = 2[cos+ isin]
zw = rr1[cos(θ +α) + isin (θ +α)]
= 2*2 [cos + isin
=4[cos + isin
= [[cos + isin
=cos - + isin
=cos - isin
=r[cosθ - isinθ]= 2[cos - isin]
b) z = +4i and w =-3 -
Z =r[cosθ + isinθ]
θ = =
r = = =8
Z =8[cos + isin]
w =-3 -
w = r1 (cos α + isinα
α = = = ; since it falls within the third quadrant
r = = =3
w = 3 [cos+ isin]
zw = rr1[cos(θ +α) + isin (θ +α)]
= 8* 3 [cos + isin
=24 [cos + isin
= [[cos + isin
= [[cos + isin]
=[cos + isin]
=r[cosθ - isinθ]= 8[cos - isin]
4. show that
a) = +
Let Z1 = a +bi and Z2 = x + yi
Z1 = a -bi and Z2 = x - yi
= . =
=
b) = .
Let Z1 = (r, θ) = r[cosθ +isinθ], = r[cosθ - isinθ],
And Z2 = (R, α) = R [cos α + isinα], = R [cos α - isinα]
= Rr [cos(θ +α) - isin (θ +α)
=R.r but R = and r =
=
c) =
z = r[cosθ + isinθ]
r.r[cos 2θ + isin2θ]
= r2 , but =r
= .
d) Z= r = r[cos θ + isinθ]
r[cos θ - isinθ]
But = cos θ - isinθ
r
5. a) (z +1)(2-i) = 3-4i
z +1 = = = = 2- i
b) =1
let w = 3 +4i
= = 5
= =1
=1
=1/5
= = =1/5
6. Z5 = -32
Let z =(ρ, θ) and w = (r, α) = -32
ρ5 [cos 5θ + isin 5θ] = r [cos α + isinα]
ρ= = 2, α = π
θ =
for other solutions of z, θ = + k2π/5
k =0
z = 2{cos }
k =1 θ = + =
z = 2{cos }
k=2, θ = + =
z = 2{cos }= -2
k=3, θ = + =
Z= 2{cos}
K=4, θ = + =
z = 2{cos }
b) Z4 +8i =0
Z4 = -8i σ= falls within third quadrant
R = = 1.682
4θ =
θ =
k =0
z = 1.682[cos + isin
k =1
θ = + =
z= 1.682[cos + isin
k=2
θ = + =
z= 1.682[cos + isin
k= 3
θ = + =
z= 1.682[cos + isin
c) z3 = -1 +
R = = = 1.587
3θ = π - = π-π/3 =
θ =
at k =0
tz= 1.587[cos + isin
k=1
θ = + =
z= 1.587[cos isin
k=2
θ = + =
z= 1.587[cos isin
7. Z2 –iz + (1 +3i) = 0
The sum of numbers = -i
Product = (1 +3i)
Nos. (1-2i) and (-1 + i)
Z = -(1-2i)
and z = -(-1+i)
=-1 + 2i
= 1-i
The roots of z are (-1 +2i) and (1-i)
Sequence and series tutorial solutions
1.
a) {2, 5, 8, 11} an = 2 + 3(n-1) , where n= 1,2, 3,
b) {1/2, ¼, 1/6} aa = 1/(2n), where n= 1,2, 3,
c) {-1, -1/3, 3/5, -5/6} an = (-1)n
d) {1, 0, 1, 0, 1} an =
e) {0, ½, 0, ½, 0} an =
2.
a) {n-} = ∞
Thus is the limit doess not exist and its divergent
b)
= =
the function converges to zero at infinity
c) = = = = ∞ ,
Since
The limit does not exist thus it is divergent.
d) Sin (+ ) = = 1
Since
+++++++++++++
Since
++++++
e) = = ∞
Since
++++++++++++++
Since
++++++++++++++*+++
f) = = . = 1 . 1 = 1
Since
+++++++++++++++
Since
+++++++++++++++*+++++++
g) = =
=0
Since
++++++++++++++++
Since
+++++++
h) =
= ln π =0
= = 1
Since
+++++++++++++++++
Since
++++++++++++
i) + = ln π += 0
Since
++++++++++++++++++
Since
++++++++++++++
3.
a) 5 + 3 + + + +
Since
+++++++++++++++++++
Since
+++++++
= 3/5 < 1 thus the series converges.
Since
++++++++++++++++++++*+++
b) 2 + 0.4 + 0.08 + 0.016 + +
Since
+++++++++++++++++++++
Since
the series converge ; since = 0.2 < 1 the sum = = = = c) 1/9 -1/3 + 1- 3 + (-1)(n+1)a(3)n-1
Since
++++++++++++++++++++++
Since
+++++++
d) + + + +
Since
+++++++++++++++++++++++*+
e)
Since
++++++++++++++++++++++++
Since
++++++++++++++++++++++++
The sum = 0
f) = 1/5 + 1/50 + 1/500 + 1/5000 + + a(1/10)n-1
Since
+++++++++++++++++++++++++
Since
++++++++
The series converges since = 1/10 < 1
The sum = = =
g)
Since
++++++++++++++++++++++
Sk= - }+{ } + + + + + +
Sk = ½ + ¼ = ¾
= sk = ¾
4)
a)
Since
++++++++++++++++
And = thus the series is divergent.
b)
Since
++++++++++++++++++++++++++++
Since
++++
Therefore; = ∞
Thus the series is divergent
c)
Since
++++++++++
= 0 an converges, so that =0
Since
++++++++++++++++++++++++++++++*++++
d)
Since
+++++++++++++++++++++++++++++
e)
Since
+++++++++++++++++++++++++++++
5. 1 +2r + r2 +2r3 +r4 + 2r5+ r6 +
= {1 + r2+r4+ r6 + + r2n} +{2r + +2r3 + 2r5 ++ 2r2n+1}
Since
+++
if the series converge
-1
Since
+++++++++++++++++
-1
Since
++++++++++++++++++++++++++
6. determine whether the series is absolutely convergent, conditionally convergent or divergent
a)
= the series is absolutely convergent.
b) -
= - = 1-1 =0
Since
+++++++++++++++++++++++++++++++++
c)
= . = 0
Since
+++++++++++++++++++++++++++++++++
d)
= = 1 test inclusive
Since
++++++++++++++++++
= {ln (x +5)} = the series is divergent
7. Find the radius of convergence and interval of convergence of the power aeries
a)
= = . .=
Since
+++++++++++++++++++
2
I (-8, 2)
b)
=. =0 < 1
Since
++++++++++++++++++++++++++++++++++++++++
Since
++
c)
= . =
Since
+++++++++++++++++++++++
For so that I(-4, 3)
d)
= 3x. = 3x
Since
+++++++++++++++++++++++
For
Since
+++++++++
e)
= . =
Since
++++++++++++++++++
For
Since
++++++++++++
f)
= . =
Since
+++++++++++++++++++
At
Since
++++++++++++++++++++++++
g)
=
Since
++++++++++++++++++++++
At
Since
++++++++++++++++++
8. the Taylor series for the functions
a) f(x) = 3x2 + 2x + 1 at a =3
Since
++++
f1(3) = 6x+ 2 = 20
f11(3) = 6
f(x) = 34 + 20(x-3) + (x-3)2
b)
f(x) = sinx a =π/2
Since
++++++++++++++++++
g(x) =
Since
++++
h1(
Since
++++
h111(
h(x) = 1- +
f(x) = g(x)h(x)
f(x) = {[1- ]. [
f(x)=
f(x)=
c) f(x) = 3x a =1
Since
+++
f1(x) = 3x ln3 f(1) = 3 ln3
f11(x) = 3x ln3. ln3; f11(1) = 3 (ln3)2
f111(x) = 3x ln3. ln3 . ln3 ; f111(1) = 3(ln3)3
f(x)= 3 + 3 ln3(x-1) + + +
d) f(x) = a =2
Since
+++
f1(x) = , f1(2) = 1
f11(x) = f11(2) = 2
f111(x) = , f111(2) =6
f(x) = 1 + + + +
9. Maclaurin series for the following functions
a) f(x) =
Since
+++
f1(x) = f1(0) =1
f11(x) = f11(0) = 2
f111(x) = f111(0) = 6
f(x) = 1 + x + + 6 +
f(x) = 1+ x+ x2 + x3 + + xn
b)
Since
+++++++
f(0) = 1
f1(x) = -3sinx = 0 at x =0
f11(x) = -9cos x = -9 at x= 0
f111(x) = 27sin x = 0
f111(x) = 81cos x = 81
f1v(x) = -243sin x = 0
fv(x) = -729cosx = -729
f(x) = 1 -3 + 81 -729 + +
c) f(x) = x
Since
+++
f1(x) = = 1 at x =0
f11(x) = = 2
f111(x) = = 3
f1v(x) = = 4
fv(x) = = 5
f(x) = 0 + x + 2 + 3 + 4 +
f(x) = x + + + + +
