Question

1+3i/1-2i find the modulus and argument also polar form of the complex number .... pl ans me ​

Answer 1

the modulus and argument of 1+3i/1-2i is  √2 and -45° respectively.

The polar form of the complex number is √2 e^{-i π/4}

• Given,
• 1+3i/1-2i
• = 1+3i/1-2i × 1+2i/1+2i
• = (1+3i)(1+2i) / (1-2i)(1+2i)
• = -5+5i / 1^2 - (2i)^2
• = -5+5i / 5
• = -1+i
• modulus =
• ∴ r  $\sqrt{x^2+y^2}\\ \\=\sqrt{(-1)^2+1^2} \\\\=\sqrt{2}$
• Ф = tan^-1  (b/a)
• = tan^-1 (1/-1)
• = tan^-1 (-1)
• ∴ Ф = -45°
• Polar form is given by,
• z = r (cos Ф + i sin Ф)
• = √2  [cos (-45° ) + i sin (-45 ° )]
• = √2  [cos (45° ) - i sin (45 ° )]
• ∴ z= √2 e^{-i π/4}

Answer 2

the modulus and argument of 1+3i/1-2i is  √2 and -45° respectively.

The polar form of the complex number is √2 e^{-i π/4}

The polar form of a complex number is another way to represent a complex number.

The form z=a+bi is called the rectangular coordinate form of a complex number.

We usually measure θ so that it lies between -180 degrees to +180 degrees.

Angles measured anticlockwise from the positive x axis are conventionally positive, whereas angles measured clockwise are negative.

According to the question,

1+3i/1-2i

= 1+3i/1-2i × 1+2i/1+2i

= (1+3i)(1+2i) / (1-2i)(1+2i)

= -5+5i / 1^2 - (2i)^2

= -5+5i / 5

= -1+i

modulus =

∴ r  

Ф = tan^-1  (b/a)

= tan^-1 (1/-1)

= tan^-1 (-1)

∴ Ф = -45°

Polar form is given by,

z = r (cos Ф + i sin Ф)

= √2  [cos (-45° ) + i sin (-45 ° )]

= √2  [cos (45° ) - i sin (45 ° )]

∴ z= √2 e^{-i π/4}