Question

Find the modulus of the arguments of the complex number 1+3i÷1-2i and convert it into polar form

Answer 1

$\frac{1 + 3i}{1 - 2i} \\ rationalize \\ \frac{(1 + 3i)(1 + 2i)}{1 + 4} \\ = \frac{ - 5+ 5i}{5} \\ = - 1 + i \\ modulus = \sqrt{1 {}^{2} + {1}^{2} } = \sqrt{2} \\ argument \\ \alpha = {tan}^{ - 1} ( - 1) = - \frac{\pi}{4}$
polar form
$= r cis \alpha = {e}^{ - i\frac{\pi}{4} } \\ = r \cos( - \frac{\pi}{4} ) + i \sin( - \frac{\pi}{4} )$
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r=√2

Answer 2

The give eqn is 1+3i ÷ 1-2i
rationalise the given equation you will get
-1 + i
now calculate the modulus
|-1 + i| = √[(-1)^2 + (1)^2]
= √(2)
for arguments we have to use the polar form that is rcosø + irsinø = z

so √(2)cosø + i√(2)sinø = -1 + i

on comparing the real value and imaginary value we have

√(2)cosø = -1
cosø = -1/√(2)

√(2)sinø = 1
sinø = 1/√(2)

so
ø = π + π/4
ø = 5π/4. [in this the value of sin is +ve and cos is -ve the ø should be in Ist quadrant]
so modulus and arguments are
√(2) and 5π/4 respectively.

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