x2 t01 04 mod arg form(2013)

Mod-Arg Form

Mod-Arg Form

y

xO

ModulusThe modulus of a complex number is the length of the vector OZz = x + iy

x

y

Mod-Arg Form

y

xO


x

y 22

222

yxr

yxr

zr

Mod-Arg Form

y

xO


x

y 22

222

yxr

yxr

zr

22 yxz

Mod-Arg Form

y

xO


x

y 22

222

yxr

yxr

zr

22 yxz

ArgumentThe argument of a complex number is the angle the vector OZ makes with the positive real (x) axis

zarg

Mod-Arg Form

y

xO


x

y 22

222

yxr

yxr

zr

22 yxz

ArgumentThe argument of a complex number is the angle the vector OZ makes with the positive real (x) axis

zarg

xyz 1tanarg zarg

ige 44ofargument andmodulus theFind ..

22 4444 i


2432

22 4444 i


44tan44arg 1i

2432

1tan 1

22 4444 i


44tan44arg 1i

2432

1tan 1

4

22 4444 i


44tan44arg 1i

2432

1tan 1

4

Every complex number can be written in terms of its modulus and argument

22 4444 i

iyxz


44tan44arg 1i

2432

1tan 1

4


sincos irr sincos ir

22 4444 i

iyxz

zzrarg

where;


44tan44arg 1i

2432

1tan 1

4


sincos irr sincos ir

The mod-arg form of z is; sincos irz

rcisz

iie.g. 44

iie.g. 44

424 cis

iie.g. 44

424 cis

iii 3

iie.g. 44

424 cis

iii 3

24

13 22

z

iie.g. 44

424 cis

iii 3

24

13 22

z3

1tanarg 1z

6

iie.g. 44

424 cis

iii 3

24

13 22

z3

1tanarg 1z

6

623 cisi

iie.g. 44

424 cis

iii 3

24

13 22

z3

1tanarg 1z

6

623 cisi

Convert 6cis to Cartsian form 6

ii

iie.g. 44

424 cis

iii 3

24

13 22

z3

1tanarg 1z

6

623 cisi


ii

6cis 6(cos sin )6 6 6

i

iie.g. 44

424 cis

iii 3

24

13 22

z3

1tanarg 1z

6

623 cisi


ii

6cis 6(cos sin )6 6 6

i

3 16( )2 2

3 3 3

i

i

Mod-Arg Relations

Mod-Arg Relations 2121 1 zzzz

2121 argargarg zzzz


2121 argargarg zzzz

NOTE: Multiplication rotates z1 by arg z2


222111 andlet cisrzcisrz

2121 argargarg zzzz

22211121 sincossincos irirzz Proof:




2121 argargarg zzzz


2121212121 sinsinsincoscossincoscos iirr 2121212121 sincoscossinsinsincoscos irr




2121 argargarg zzzz



212121 sincos irr




2121 argargarg zzzz

22211121 sincossincos irirzz

21

2121

zzrrzz

Proof:


212121 sincos irr




2121 argargarg zzzz


21

2121

zzrrzz

Proof:


212121 sincos irr

21

2121

argarg arg

zzzz




2121 argargarg zzzz


21

2121

zzrrzz

Proof:


212121 sincos irr

21

2121

argarg arg

zzzz

NOTE: it follows that;nn zzzzzzzz 321321

nn zzzzzzzz argargargargarg 321321


2

1

2

1 2zz

zz

212

1 argargarg zzzz

2

1

2

1 2zz

zz

212

1 argargarg zzzz

NOTE: it follows that;

43

21

43

21

zzzz

zzzz

432143

21 argargargargarg zzzzzzzz

2

1

2

1 2zz

zz

212

1 argargarg zzzz

NOTE: it follows that;

43

21

43

21

zzzz

zzzz

432143

21 argargargargarg zzzzzzzz

nn zz 3

znzn argarg

i

iize.g.2325 ofargument and modulus theFind

i


22

2222

231215

z

i


22

2222

231215

z

13526

10

i


22

2222

231215

z

13526

10

32tan

21tan

51tanarg 111z

i


22

2222

231215

z

13526

10

32tan

21tan

51tanarg 111z

84175

1433621539111

i


22

2222

231215

z

13526

10

32tan

21tan

51tanarg 111z

84175

1433621539111

Patel: Exercise 4B; evens

Patel: Exercise 4C; 1 to 10 evens

Cambridge: Exercise 1D; 9, 11, 13, 14, 16, 20

x2 t01 04 mod arg form(2013)

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