Question

Show that z=5÷(1-i)(2-i)(3-i) is purely imaginary number

Answer 1

Answer:  The proof is done below.

Step-by-step explanation:  We are given to show that the following complex expression is purely imaginary :

$z=5\div(1-i)(2-i)(3-i)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We know that

a complex expression is purely imaginary if the real part is zero.

We will be using the following values of powers of imaginary number i :

$i=\sqrt{-1},~~~~i^2=-1.$

From (i), we have

$z\\\\=5\div(1-i)(2-i)(3-i)\\\\\\=\dfrac{5}{(1-i)(2-i)(3-i)}\\\\\\=\dfrac{5}{(2-i-2i+i^2)(3-i)}\\\\\\=\dfrac{5}{(2-3i-1)(3-i)}\\\\\\=\dfrac{5}{(1-3i)(3-i)}\\\\\\=\dfrac{5}{3-i-9i+3i^2}\\\\\\=\dfrac{5}{3-10i-3}\\\\\\=\dfrac{5}{-10i}\\\\\\=\dfrac{i^2}{2i}\\\\\\=\dfrac{i}{2}.$

So, there is no real part in the simplified value of the given expression.

Thus, z is a purely imaginary number.

Hence showed.

Answer 2

Step-by-step explanation:

the above image has the answer