SOLUTION: Divide: 1-4i/5-6i. Write your answer in a+bi form A= B= I'm really confused on the a+bi thing, as well as solving the fraction. I tried to cancel out the i's, but I think that's

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Divide: 1-4i/5-6i. Write your answer in a+bi form A= B= I'm really confused on the a+bi thing, as well as solving the fraction. I tried to cancel out the i's, but I think that's      Log On


   

Question 1112642: Divide: 1-4i/5-6i. Write your answer in a+bi form
A=
B=
I'm really confused on the a+bi thing, as well as solving the fraction. I tried to cancel out the i's, but I think that's the opposite of what I'm supposed to do.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
Best also to write the expression correctly.

%281-4i%29%2F%285-6i%29

You want to simplify this and you use the conjugate of the denominator; and multiply the given expression by 1.

%28%281-4i%29%2F%285-6i%29%29%28%285%2B6i%29%2F%285%2B6i%29%29

Do you know the next step(s) ?
Understand that i%2Ai=-1.
-

Do the multiplication indicated and simplify from there if possible.
-
highlight%2829%2F61-%2814%2F61%29i%29

Answer by ikleyn(52752) About Me  (Show Source):
You can put this solution on YOUR website!
.
  %281-4i%29%2F%285-6i%29 = 


It is standard problem on complex numbers. I will show you the standard way of solving it.


Multiply the numerator and the denominator by the complex number conjugate to that of denominator.

In your case the denominator is  5-6i  and  the conjugate number to it  is  5+6i.

So, we will multiply the numerator and denominator by (5+6i).


    Since we multiply the numerator and denominator by the same number, the value

    of our fraction remains the same.  So, we can continue from the above 


= %281-4i%29%2F%285-6i%29 . %285%2B6i%29%2F%285%2B6i%29 = %28%281-4i%29%2A%285%2B6i%29%29%2F%28%285-6i%29%2A%285%2B6i%29%29


Now it is better (until you gain the necessary practice) to work separately with the numerator and denominator.


Numerator = (1-4i)*(5+6i) = 5 - 20i + 6i - 24*(i^2) = 5 + 24 - 14i = 29-14i    (((<<<---=== you remember, of course, that  i^2 = -1)


Denominator = (5-6i)*(5+6i) = 25 - 30i + 30i - 36*(i^2) = 25+36 = 61.    


    The modified denominator is the product of the complex number and its conjugate, so it is a REAL NUMBER.  

    It is why we multiplied by the conjugate number:  to get a real number in the denominator !


So, our fraction is  NUM%2FDEN = %2829-14i%29%2F61 = 29%2F61 - %2814%2F61%29%2Ai.


It is your final presentation of the given fraction as a complex number in the form  z = a + bi.


In your case  a = 29%2F61,  b = -14%2F61.

--------------
Completed and solved.

What I showed to you is the standard method solving such problems.

Memorize it as a mantra.


================

There is a bunch of introductory lessons on complex numbers
    - Complex numbers and arithmetical operations on them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
    - Raising a complex number to an integer power
    - How to take a root of a complex number

    - Solved problems on taking roots of complex numbers
    - Solved problems on arithmetic operations on complex numbers
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Complex numbers".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.