Question

What is the product of the complex numbers 3-4i3−4i3, minus, 4, i and -6+i−6+iminus, 6, plus, i, where i=\sqrt{-1}i= −1 ​ i, equals, square root of, minus, 1, end square root ?

Answer 1

Answer:

$z_1\times z_2=-14+27i$

Step-by-step explanation:

We are given that two complex numbers

$z_1=3-4i$

$z_2=-6+i$

We have to find the product of two complex numbers

$z_1\times z_2=(3-4i)(-6+i)$

$z_1\times z_2=3(-6+i)-4i(-6+i)$

$z_1\times z_2=-18+3i+24i-4i^2$

$z_1\times z_2=-18+27i+4$ ($i^2=-1$)

$z_1\times z_2=-14+27i$