Notes Complex Numbers Exponential form IBHL

Another form to write a complex number in is Euler’s Form (Exponential Form):

z=r e iθ (this is sometimes written as z=r e i(θ+2kπ ), k ϵ Z)

To show where this comes from we need to use Euler’s formula: e iθ=cosθ+i sin θTherefore: z=r ¿

z=r e iθ

Find the following:

z1 z2=¿

z1 z2=r 1r2ei θ1 eiθ2=r1 r2 e

i (θ1+θ¿¿2)¿

z1z2

=¿

z1z2

=r1 e

i θ1

r2 ei θ2

=r1r2e i(θ1−θ¿¿2)¿

zn=¿

zn=( re iθ )n=r ne inθ

Calculate (3+3 i)5.

Solution: We could first write this in exponential form.

r=√32+32=3√2, tanθ=33 ,∴θ=π4

Therefore (3+3 i )5=3√2 ei π4 .

(3+3 i )5=(3√2 )5(ei π4 )5

=972√2ei 5π4

Then we can convert this back:

972√2ei 5π4 =972√2cos ( 5π4 )+i 972√2sin( 5 π4 )

1

Notes Complex Numbers Exponential form IBHL

¿972√2(−√22 )+i 972√2(−√2

2 )

¿−972−972i

Nth Roots of Unity: ( z )n=1

z=1 is the obvious solution but are there more:

(r e iθ )n=1 ei0 (since 1=ei0)

(r )ne inθ=1 ei0

Therefore (r )n=1 and nθ=0+2 πk, where k∈ {0 ,±1 ,±2 , ±3 ,… }

So r=1, θ=2 πkn , k∈ {0 ,±1 ,±2 , ±3 ,… }

z=ei 2πkn this is sometimes written as z=exp (i 2πkn )

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Notes Complex Numbers Exponential form IBHL

Complex root in Euler form:

3