Question

find the argument of 1 +√ 3 i​

Answer 1

Answer:

Let z=

1+i

3

1−i

3

⇒z=

1+i

3

1−i

3

×

1−i

3

1−i

3

⇒z=

1+3

(1−i

3

)

2

⇒z=

4

1−3−2i

3

⇒z=−

2

1

−

2

i

3

r=

(−

2

1

)

2

+(−

2

3

)

2

=1

Comparing the above equation with z=rcosα+irsinα

rcosα=−

2

1

⇒cosα=−

2

1

rsinα=−

2

3

⇒sinα=−

2

3

Since sinα and cosα, both are negative, thus the argument will be in III

rd

quadrant.

α=180°+60°=240°(∵sin60°=

2

3

&cos60°=

2

1

)

Hence argument of given complex is 240°.