Question #220737

Let z1 =-i/1+i,z2 = 1+i/1-i,and z3 = 1/10[2(i-1)1 +(i +√3)3+(1- i)(1-i)].Express z1z3/z2,z1z2/z3,and z1/z3z2 in both polar and standard forms.


1
Expert's answer
2021-07-27T16:22:57-0400
z1=i1+i,z2=1+i1i,z_1=-\dfrac{i}{1+i}, z_2=\dfrac{1+i}{1-i},

z3=110[2(i1)(1i)+(i+3)3+(1i)(1i)]z_3=\dfrac{1}{10}[2(i-1)(1-i) +(i +√3)^3+(1- i)(1-i)]

z1=i1+i=i(1i)2=1212iz_1=-\dfrac{i}{1+i}=-\dfrac{i(1-i)}{2}=-\dfrac{1}{2}-\dfrac{1}{2}i

z2=1+i1i=(1+i)22=iz_2=\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{2}=i

(i+3)3=i33+9i+33=8i(i+\sqrt{3})^3=-i-3\sqrt{3}+9i+3\sqrt{3}=8i

(1i)(1i)=12i1=2i(1-i)(1-i)=1-2i-1=-2i

2(i1)(1i)=4i2(i-1)(1-i)=4i

z3=110[4i+8i2i]=iz_3=\dfrac{1}{10}[4i+8i-2i]=i

z1z3z2=(1212i)ii=1212i\dfrac{z_1z_3}{z_2}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)i}{i}=-\dfrac{1}{2}-\dfrac{1}{2}i

=22(cos(3π4)+isin(3π4))=\dfrac{\sqrt{2}}{2}(\cos(-\dfrac{3\pi}{4})+i\sin(-\dfrac{3\pi}{4}))


z1z2z3=(1212i)ii=1212i\dfrac{z_1z_2}{z_3}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)i}{i}=-\dfrac{1}{2}-\dfrac{1}{2}i

=22(cos(3π4)+isin(3π4))=\dfrac{\sqrt{2}}{2}(\cos(-\dfrac{3\pi}{4})+i\sin(-\dfrac{3\pi}{4}))


z1z2z3=(1212i)(i)i=12+12i\dfrac{z_1}{z_2z_3}=\dfrac{(-\dfrac{1}{2}-\dfrac{1}{2}i)}{(i)i}=\dfrac{1}{2}+\dfrac{1}{2}i

=22(cos(π4)+isin(π4))=\dfrac{\sqrt{2}}{2}(\cos(\dfrac{\pi}{4})+i\sin(\dfrac{\pi}{4}))



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