1,057
14
Essay, 6 pages (1400 words)

Tutorial solutions complex numbers and sequence essay sample

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 < 1the 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 < 1The 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) = 3×2 + 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 + + + + +

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