The modulus-amplitude form of ((1-i)3(2-i)/(2+i)(1+i)) is

Question

The modulus-amplitude form of ((1-i)3(2-i)/(2+i)(1+i)) is (A) 2 textcis(π- tan -1 (4/3)) (B) 2 textcis(- tan -1 (4/3)) (C) 2 textcis(-π+ tan -1 (4/3)) (D) 2 textcis( tan -1 (4/3))

Tardigrade Admin · Accepted Answer

We have, ((1-i)3(2-i)/(2+i)(1+i))=((1-i)2(1-i)(2-i)/(1+i)(2+i)) =(-2 i(1-i)(1-i)(2-i)(2-i)/(1+i)(1-i)(2+i)(2-i)) =(-2 i(-2 i)(3-4 i)/2 ⋅ 5)=-2((3/5)-(4 i/5)) =2((-3/5)+(4/5) i) =2 cis(π- tan -1((4/3)))

Q. The modulus-amplitude form of $\frac{( 1 - i ) ^{3} ( 2 - i )}{( 2 + i ) ( 1 + i )}$ is

$2 cis (π - tan^{- 1} \frac{4}{3})$

$2 cis (- tan^{- 1} \frac{4}{3})$

$2 cis (- π + tan^{- 1} \frac{4}{3})$

$2 cis (tan^{- 1} \frac{4}{3})$

Q. The modulus-amplitude form of (2+i)(1+i)(1−i)3(2−i)​ is

2cis(π−tan−134​)

2cis(−tan−134​)

2cis(−π+tan−134​)

2cis(tan−134​)

We have, (2+i)(1+i)(1−i)3(2−i)​=(1+i)(2+i)(1−i)2(1−i)(2−i)​ =(1+i)(1−i)(2+i)(2−i)−2i(1−i)(1−i)(2−i)(2−i)​ =2⋅5−2i(−2i)(3−4i)​=−2(53​−54i​) =2(5−3​+54​i) =2cis(π−tan−1(34​))

Q. The modulus-amplitude form of $\frac{( 1 - i ) ^{3} ( 2 - i )}{( 2 + i ) ( 1 + i )}$ is

$2 cis (π - tan^{- 1} \frac{4}{3})$

$2 cis (- tan^{- 1} \frac{4}{3})$

$2 cis (- π + tan^{- 1} \frac{4}{3})$

$2 cis (tan^{- 1} \frac{4}{3})$