Question 742206
.
4-i over
2-6i
can you walk me through the steps Im a little rusty.
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<pre>
You are given a fraction of complex numbers  {{{(4-i)/(2-6i)}}}.

They want you transform it to the standard form complex number.


To do it, multiply the numerator and denominator by the complex number,
which is conjugate to the denominator of the given fraction.

So, you write

    {{{(4-i)/(2-6i)}}} = {{{((4-i)/(2-6i))*((2+6i)/(2+6i))}}} = {{{((4-i)*(2+6i))/((2-6i)*(2+6i))}}}.


Now, to avoid a mess, let's work with the numerator and the denominator separately.


In the numerator, you have

    (4-i)*(2+6i) = 8 -2i + 24i - 6i^2 = 8 + 22i - 6*(-1) = 8 + 6 + 22i = 14 + 22i.


In the denominator, you have the product of a complex number (2-6i) and its conjugate (2+6i),
which is a real number

    (2-6i)*(2+6i) = 2^2 + 6^2 = 4 + 36 = 40.


Thus, the ratio  {{{(4-i)/(2-6i)}}}  is

    {{{(4-i)/(2-6i)}}} = {{{(14+22i)/40}}} = {{{7/20 + (11/20)i}}}.


It is a standard form of a complex number.


At this point, the assignment is complete.


<U>ANSWER</U>.  {{{(4-i)/(2-6i)}}} = {{{7/20 + (11/20)*i}}}.

         Or, which is the same,  {{{(4-i)/(2-6i)}}} = 0.35 + 0.55*i.
</pre>

Solved.


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The answer in the post by @lynnlo is incorrect.