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Mathematics Question on Complex Numbers and Quadratic Equations

Let z1=2–i,z2=–2+i. Find (i)Re(z1z2z1),(ii) Im(1z1z1)Let\, z_1=2 – i,z_2 = –2+i.\, Find\, (i) Re(\frac{z_1z_2}{z_1}), (ii) \,Im(\frac{1}{z_1z_1})

Answer

z!=2−i,z@=−2+iz_!=2-i,z_@=-2+i

(i)z1z2=(2−i)(−2+i)=−4+2i+2i−i2=−4+4i−(−1)=−3+4i(i)z_1z_2=(2-i)(-2+i)=-4+2i+2i-i^2=-4+4i-(-1)=-3+4i

z1ˉ=2+1\bar{z_1}=2+1

∴z1z2z1=−3+4i2+i∴\frac{z_1z_2}{z_1}=\frac{-3+4i}{2+i}

On multiplying numerator and denominator by (2i), we obtain

z1z2z1=(−3+4i)(2−i)(2+i)(2−i)=−6+3i+8i−4i222+12=−6+1li−4(−1)22+12\frac{z_1z_2}{z_1}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i-4i^2}{2^2+1^2}=\frac{-6+1li-4(-1)}{2^2+1^2}

=−2+1 li5=−25+115i=\frac{-2+1\,li}{5}=\frac{-2}{5}+\frac{11}{5}i

On comparing real parts, we obtain

Re(z1z2z1ˉ)=−25Re(\frac{z_1z_2}{\bar{z_1}})=\frac{-2}{5}

(ii) 1z1z1ˉ=1(2−i)(2+i)=1(2)2+(1)2=15\frac{1}{z_1\bar{z_1}}=\frac{1}{(2-i)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}

On comparing imaginary parts, we obtain

Im(1z1z1ˉ)=0Im(\frac{1}{z_1\bar{z_1}})=0