Question
Download Solution PDFWhat is the modulus of \(\rm \dfrac{4+2i}{1-2i}\) where \(\rm i=\sqrt{-1} ?\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Let z = x + iy be a complex number, Where x is called real part of the complex number or Re (z) and y is called Imaginary part of the complex number or Im (z)Modulus of z = |z| = \(\rm \sqrt {x^2+y^2} = \sqrt {Re (z)^2+Im (z)^2}\)
Calculations:
Let \(\rm z= x + iy = \dfrac{4+2i}{1-2i}\)
\(\rm = \dfrac{4+2i}{1-2i}\times\dfrac{1+2i}{1+2i}\)
\(\rm= \dfrac{4+10i+4i^2}{1-4i^2}\)
As we know i2 = -1
\(\rm = \dfrac{4+10i-4}{1+4}\)
\(\rm x + iy =\dfrac{10i}{5} = 0 + 2i\)
As we know that if z = x + iy be any complex number, then its modulus is given by,|z| = \(\rm \sqrt{x^2+y^2}\)
∴ |z| = \(\rm \sqrt{0^2+2^2} = 2\)
Last updated on May 31, 2025
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