(i) ∣2z1−4z2∣|2z_1-4z_2|∣2z1−4z2∣ Solution Given z1=2+7i, z2=−5+3i.2z1=2(2+7i)=4+14i4z2=4(−5+3i)=−20+12i2z1−4z2=(4+14i)−(−20+12i)=24+2i∣2z1−4z2∣=∣24+2i∣=242+22=580=2145\begin{aligned} & \boxed{\text{Given } z_1=2+7i,\ z_2=-5+3i.} \\ \\ & 2z_1=2(2+7i)=4+14i \\ & 4z_2=4(-5+3i)=-20+12i \\ \\ & 2z_1-4z_2=(4+14i)-(-20+12i)=24+2i \\ \\ & |2z_1-4z_2|=|24+2i|=\sqrt{24^2+2^2}=\sqrt{580}=2\sqrt{145} \end{aligned}Given z1=2+7i, z2=−5+3i.2z1=2(2+7i)=4+14i4z2=4(−5+3i)=−20+12i2z1−4z2=(4+14i)−(−20+12i)=24+2i∣2z1−4z2∣=∣24+2i∣=242+22=580=2145