Question

Find the complex number z such that (1-4i)z is a real number and z+6-i is an imaginary number.

Answer 1

i think than use formula a^2 -b^2 =(a-b)(a+b)
where a=1 and b= 4i will help you finding the z what will be equal (1+b)

try it please and show your work

Answer 2

@DanJS could you help with my question please

Answer 3

The same question is in Closed questions list from a bit ago
http://openstudy.com/study#/updates/57fa098ee4b06a18dc414864

Answer 4

A number is real hen it's equal to its own complex conjugate. A number is imaginary when it's equal to the negative of its complex conjugate.

Answer 5

I would let z= a+b i
then compute
(1-4i)(a+bi) = pure real
and
a+b i+6-i = pure imaginary

notice in the 2nd equation we can write the sum as
(a+6) + (b-1) i
if it is pure imaginary, then the "real part" is zero. In other words
a+6= 0
and we can "solve for a"
now multiply out the first equation, and solve for the b value that makes the imaginary part of the product zero.

Answer 6

I might as well show my strategy, set up two equations based on \(real = real^*\) and \(imaginary = - imaginary^*\) :

\[(1-4i)z=(1+4i)z^*\]\[(z+6-i)=-(z^*+6+i)\]
second equation simplifies to:
\[\frac{z+z^*}{2}=-6\]which is the real part of \(z\).

first equation rearranges to:
\[\frac{z-z^*}{2i} = 4\frac{z+z^*}{2}\]which says the imaginary part of z is 4 times the real part of z.

Solved. :P