2020-05-20T08:20:01-04:00
9. Find the moduli and principal arguments of w, z, wz and w/z, given that (a) w = 10i, z = 1 + i
3 + 2i, z = 1 − i, (c) w = , . 1 1 + i √
1
2020-05-25T20:32:18-0400
a)
w = 10 i , ∣ w ∣ = 10 , w=10i, |w|=10, w = 10 i , ∣ w ∣ = 10 , A r g ( w ) = π 2 Arg(w)={\pi \over 2} A r g ( w ) = 2 π
z = 1 + 3 i , ∣ z ∣ = ( 1 ) 2 + ( 3 ) 2 = 2 , z=1+\sqrt{3}i, |z|=\sqrt{(1)^2+(\sqrt{3})^2}=2, z = 1 + 3 i , ∣ z ∣ = ( 1 ) 2 + ( 3 ) 2 = 2 , A r g ( z ) = π 3 Arg(z)={\pi \over 3} A r g ( z ) = 3 π
∣ w z ∣ = ∣ w ∣ ∣ z ∣ = 10 ( 2 ) = 20 , |wz|=|w||z|=10(2)=20, ∣ w z ∣ = ∣ w ∣∣ z ∣ = 10 ( 2 ) = 20 , A r g ( w z ) = A r g ( w ) + A r g ( z ) = π 2 + π 3 = 5 π 6 Arg(wz)=Arg(w)+Arg(z)={\pi \over 2}+{\pi \over 3}={5\pi \over6} A r g ( w z ) = A r g ( w ) + A r g ( z ) = 2 π + 3 π = 6 5 π
∣ w z ∣ = ∣ w ∣ ∣ z ∣ = 10 2 = 5 , |{w \over z}|={|w|\over |z|}={10 \over 2}=5, ∣ z w ∣ = ∣ z ∣ ∣ w ∣ = 2 10 = 5 , A r g ( w z ) = A r g ( w ) − A r g ( z ) = π 2 − π 3 = π 6 Arg({w \over z})=Arg(w)-Arg(z)={\pi \over 2}-{\pi \over 3}={\pi \over6} A r g ( z w ) = A r g ( w ) − A r g ( z ) = 2 π − 3 π = 6 π
b)
w = 1 1 + i 3 = 1 − i 3 ( 1 ) 2 + ( 3 ) 2 = 1 2 − i 3 2 , w={1 \over 1+i\sqrt{3}}={1-i\sqrt{3} \over (1)^2+(\sqrt{3})^2}={1 \over 2}-i{\sqrt{3} \over 2}, w = 1 + i 3 1 = ( 1 ) 2 + ( 3 ) 2 1 − i 3 = 2 1 − i 2 3 ,
∣ w ∣ = ( 1 2 ) 2 + ( 3 2 ) 2 = 1 , A r g ( w ) = − π 3 |w|=\sqrt{({1 \over 2})^2+({\sqrt{3} \over 2})^2}=1, Arg(w)=-{\pi\over 3} ∣ w ∣ = ( 2 1 ) 2 + ( 2 3 ) 2 = 1 , A r g ( w ) = − 3 π
z = 1 − i , ∣ z ∣ = ( 1 ) 2 + ( − 1 ) 2 = 2 , z=1-i, |z|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}, z = 1 − i , ∣ z ∣ = ( 1 ) 2 + ( − 1 ) 2 = 2 , A r g ( z ) = − π 4 Arg(z)=-{\pi \over 4} A r g ( z ) = − 4 π
∣ w z ∣ = ∣ w ∣ ∣ z ∣ = 1 ( 2 ) = 2 , |wz|=|w||z|=1(\sqrt{2})=\sqrt{2}, ∣ w z ∣ = ∣ w ∣∣ z ∣ = 1 ( 2 ) = 2 , A r g ( w z ) = A r g ( w ) + A r g ( z ) = − π 3 − π 4 = − 7 π 12 Arg(wz)=Arg(w)+Arg(z)=-{\pi \over 3}-{\pi \over 4}=-{7\pi \over12} A r g ( w z ) = A r g ( w ) + A r g ( z ) = − 3 π − 4 π = − 12 7 π
∣ w z ∣ = ∣ w ∣ ∣ z ∣ = 1 2 = 2 2 , |{w \over z}|={|w|\over |z|}={1 \over \sqrt{2}}={\sqrt{2} \over 2}, ∣ z w ∣ = ∣ z ∣ ∣ w ∣ = 2 1 = 2 2 ,
A r g ( w z ) = A r g ( w ) − A r g ( z ) = − π 3 − ( − π 4 ) = − π 12 Arg({w \over z})=Arg(w)-Arg(z)=-{\pi \over 3}-(-{\pi \over 4})=-{\pi \over12} A r g ( z w ) = A r g ( w ) − A r g ( z ) = − 3 π − ( − 4 π ) = − 12 π
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