The modulus of ( 3 − i ) ( 1 + i 3 ) ( 2 + 2 i ) is

3 − i ( 1 + i 3 ) ( 2 + 2 i ) = 3 − i 2 + 2 3 i + 2 i − 2 3 = 3 − i ( 2 − 2 3 ) + ( 2 3 + 2 ) i × 3 + i 3 + i = 3 + 1 2 3 − 6 + 2 i − 2 3 i + 6 i + 2 3 i − 2 3 − 2 = 4 8 i − 8 = − 2 + 2 i ∴ Modulus = ( − 2 ) 2 + ( 2 ) 2 = 2 2

Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Easy2 minPYQ_2023

MathematicsEasysingle choice

The modulus of $\frac{( 1 + i 3 ) ( 2 + 2 i )}{( 3 - i )}$ is

Options:

Answer:

Solution:

$\frac{( 1 + i 3 ) ( 2 + 2 i )}{3 - i} = \frac{2 + 2 3 i + 2 i - 2 3}{3 - i} = \frac{( 2 - 2 3 ) + ( 2 3 + 2 ) i}{3 - i} \times \frac{3 + i}{3 + i} = \frac{2 3 - 6 + 2 i - 2 3 i + 6 i + 2 3 i - 2 3 - 2}{3 + 1} = \frac{8 i - 8}{4} = - 2 + 2 i ∴ Modulus = (- 2)^{2} + (2)^{2} = 22$

Stream:BITSATSubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2023

Doubts & Discussion

Loading discussions...