Find the modulus and argument of the complex number 
Let us assume, ![]()
Now multiplying the numerator and denominator by 1 + 3i, we get:
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= ![]()
= ![]()
= ![]()
= ![]()
Let us now assume, ![]()
∴ ![]()
Now, by squaring the both sides we get:
![]()
![]()
As conventionally r > 0, thus ![]()
∴ ![]()
![]()
Hence, ![]()
Thus, modulus of the given complex number = ![]()
And, argument of given complex number = ![]()
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