Question

find the modules are argument of complex numbers 1 by1+1​

Answer 1

Answer:

Let, z=

1−3i

1+2i

z=

1−3i

1+2i

×

1+3i

1+3i

=

1

2

+3

2

1+3i+2i+6i

2

=

1+9

1+5i+6(−1)

=

10

−5+5i

=

2

−1

+

2

1

i

Let z=rcosθ+irsinθ

i.e.,rcosθ=

2

−1

and rsinθ=

2

1

On squaring and adding, we obtain

r

2

(cos

2

θ+sin

2

θ)=(

2

−1

)

2

+(

2

1

)

2

⇒r

2

=

4

1

+

4

1

=

2

1

As r>0

⇒r=

2

1

∴

2

1

cosθ=

2

−1

and

2

1

sinθ=

2

1

⇒cosθ=

2

−1

and sinθ=

2

1

∴θ=π−

4

π

=

4

3π

[As θ lies in the II quadrant ]

Therefore the modulus and argument of the given complex number are

2

1

and

4

3π

respectively.