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Question Number 99623 by hardylanes last updated on 22/Jun/20

obtain the modulus and arguement of (((1−i)^4 )/((2+2(√(3i)^3 ))))

obtainthemodulusandarguementof(1−i)4(2+23i)3

Commented by Dwaipayan Shikari last updated on 22/Jun/20

=(((1−i)^(2.2) )/(4^3 ((1/2)+((√3)/2)i)^3 ))=(((−2i)^2 )/(64e^(((πi)/3).3) ))=((−4)/(64 e^(πi) ))=((−4)/(−64))=(1/(16))+0.i=z mod (z) is (1/(16)) arg(z) is 0 [As α=tan^(−1) ((0/(1/(16))))=0 ]{And e^(πi) =−1}

=(1−i)2.243(12+32i)3=(−2i)264eπi3.3=−464eπi=−4−64=116+0.i=zmod(z)is116arg(z)is0[Asα=tan−1(0116)=0]{Andeπi=−1}

Answered by Rio Michael last updated on 22/Jun/20

let z = (((1−i)^4 )/((2 + 2(√3) i)^3 )) = ((((√2) e^(i((7π)/4)) )^4 )/((4e^(i(π/3)) )^3 )) = ((4e^(7πi) )/(4^3 e^(πi) )) ⇒ ∣z∣ = (4/4^3 ) = (1/4^2 ) = (1/(16)) arg z = 7π−π = 6π = 3(2π) aco−terminal angle = 0 ⇒ arg z = 0

letz=(1−i)4(2+23i)3=(2ei7π4)4(4eiπ3)3=4e7πi43eπi⇒∣z∣=443=142=116argz=7π−π=6π=3(2π)aco−terminalangle=0⇒argz=0

Commented by mathmax by abdo last updated on 23/Jun/20

z =(((1−i)^4 )/((2+2(√3)i)^3 )) we have 1−i =(√2)e^(−((iπ)/4)) ⇒(1−i)^4 =4 e^(−iπ) 2+2(√3)i =2(1+(√3)i) =4((1/2)+((i(√3))/2)) =4e^((iπ)/3) ⇒(2+2(√3)i)^3 =4^3 e^((4iπ)/3) ⇒ z =(1/(16)) e^(−iπ−((4iπ)/3)) =(1/(16)) e^(−i(π+((4π)/3))) =(1/(16)) e^(−i(((7π)/3))) =(1/(16)) e^(−i(2π +(π/3))) =(1/(16)) e^(−((iπ)/3)) ⇒ ∣z∣ =(1/(16)) and argz ≡−(π/3)[2π]

z=(1−i)4(2+23i)3wehave1−i=2e−iπ4⇒(1−i)4=4e−iπ2+23i=2(1+3i)=4(12+i32)=4eiπ3⇒(2+23i)3=43e4iπ3⇒z=116e−iπ−4iπ3=116e−i(π+4π3)=116e−i(7π3)=116e−i(2π+π3)=116e−iπ3⇒∣z∣=116andargz≡−π3[2π]

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