the complex number z=5/(1-i) (2-i) (3-i)
Answers
Answer:
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)z=5÷(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.i=
−1
, i
2
=−1.
From (i), we have
\begin{gathered}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}.\end{gathered}
z
=5÷(1−i)(2−i)(3−i)
=
(1−i)(2−i)(3−i)
5
=
(2−i−2i+i
2
)(3−i)
5
=
(2−3i−1)(3−i)
5
=
(1−3i)(3−i)
5
=
3−i−9i+3i
2
5
=
3−10i−3
5
=
−10i
5
=
2i
i
2
=
2
i
.
So, there is no real part in the simplified value of the given expression.
Thus, z is a purely imaginary number.