Find modulus and argument of 1 2i 1 3i

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

Answered by admin @


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}

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