Question

(1+2/i)(3+4/i)(5+I)^-1​

Answer 1

Answer:

Heya mate your answer :)

Step-by-step explanation:

z=(11−4i−21+i)(3−4i5+i)

=(1+i−2+8i(1−4i)(1+i))(3−4i5+i)

=(−1+9i1+i−4i+4)(3−4i5+i)

=(−1+9i5−3i)(3−4i5+i)

=(−1+9i)(3−4i)(5−3i)(5+i)

=−3+4i+27i−36i225+5i−15i−3i2

=−3+4i+27i+3625+5i−15i+3

=33+31i28−10i∗28+10i28+10i

=924+868i+330+310i2784−100i2

=924+868i+330i−310784+100

=614+1198i884

z=307442+599442i