Complex numbers wk 7

Question 1

Complex numbers can be represented as…

Answer 1

points in a two-dimensional plane which is called the Argand
diagram or the complex plane.

Question 2

what can j be written as

Answer 2

Question 3

notation for complex no. form

Answer 3

z= a +bj

real part of z = Re(z)=a

imaginary part of z = lm(z)= b

Question 4

complex conjugate of z=a+bj

Answer 4

also z* is used

Question 5

find complex conjugate of these:

Answer 5

NOTE!!
Real number stays same
but imaginary part (with j), has opposite sign

Question 6

notation of complex no. form equations (2)

Answer 6

Re(z) = 1/2 (z + zbar)
Re(z) = a (real part)

lm(z) = -1/2(j)(z-zbar)
lm(z) = b (imaginary part)

Question 7

properties of conjugation (4)

Answer 7

1)________ _ _
z1 + z2 = z1 + z2

2) _______ _ _
z1x z2 = z1 + z2

3) _______ __ __
(z1/z2) = (z1)/ (z2)

4) _
zz = a^2 + b^2

Question 8

solve

Answer 8

Question 9

solve

Answer 9

j^2= -1

Question 10

Answer 10

1) find conjugate of no.
(real part stays same, imaginary part has opposite sign)

2) simplify denominator/numerator
j^2= -1

3) simplify the fraction

Question 11

complex solutions always occur in..

Answer 11

complex conjugate pairs
e.g. has +/- solutions
like in quadratic formula when ANS has +/-

Question 12

Answer 12

Question 13

State quadratic equation solution type if pendulum is:

heavily damped

lightly damped

Answer 13

H:will not oscillate so has real solutions

L: will oscillate so has complex conjugate pair of solutions

Question 14

modulus of z ?
arguement of z?
notations

Answer 14

argument in range -pi < theta <= pi

Question 15

inverse tan quadrants +rules

Answer 15

2nd= arg z = tan^-1 (b/a) + pi

3rd= arg z = tan^-1 (b/a) - pi

1st/4th= same = arg z = tan^-1 (b/a)

Question 16

find r and theta

Answer 16

Question 17

a

Answer 17

Question 18

5 points

Answer 18

• Fundamental theorem of algebra states polynomial of degree n has exactly n solutions, which is real or complex.

Where complex solutions come in complex conjugate pairs

• so cubic equation must have 3 solutions and since complex solutions come in pairs there’s 2 possibilities:

1) all 3 solutions real
2) one solution is real and other 2 are complex conjugate pair
both cases theres at least 1 solution

Question 19

a

Answer 19

Question 20

b

Question 21

c

Question 22

cos and sin definitions in terms of r
thus express complex no. in terms of r and @

Answer 22

cos @ = a/r

sin @ = b/r

z = a + bj
= r (cos@ + jsin@)
polar form of a complex number

Question 23

multiplication and division rules for polar form

Answer 23

Question 24

properties of modulus for polar form (3)

Answer 24

Question 25

properties of argument for polar form

Answer 25

Question 26

Answer 26

Question 27

express in form a + bj

Answer 27

use Euler’s formula

=

Question 28

express in form a + bj

Answer 28

conjugate version
so real stays same and imaginary part has sign reversed

Question 29

express in form a + bj

Answer 29

use Euler’s formula

Question 30

exponential form of a complex number

Answer 30

take polar form of a complex number:
z= r (cos@ + jsin@)

rewrite using eulers formula

Question 31

find imaginary number and real number from this

Answer 31

real number= e^a

imaginary= b

Question 32

Answer 32

1) break powers into 2 (one with j components)

2) anything outside the j = @

use e^j@= cos@ + jsin@

3) sub in @

4) write real values and simplify

Question 33

as a + bj

Answer 33

Question 34

Answer 34

Question 35

standard form for these

Answer 35