SOLUTION: evaluate |1+2i/2+3i|

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Question 1100156: evaluate |1+2i/2+3i|
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2B2i%29%2F%282%2B3i%29

%28%281%2B2i%29%2F%282%2B3i%29%29%28%282-3i%29%2F%282-3i%29%29

%28%281%2B2i%29%282-3i%29%29%2F%284%2B9%29

%282-3i%2B4i%2B6%29%2F13

%28i%2B8%29%2F13

The last part is to find the length of this complex number, which you can do using Pythagorean Theorem formula.




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fixedmistake

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
  %281%2B2i%29%2F%282%2B3i%29 = 


= %28%281%2B2i%29%2F%282%2B3i%29%29%28%282-3i%29%2F%282-3i%29%29 =


= %28%281%2B2i%29%282-3i%29%29%2F%284%2B9%29 =


= %282-3i%2B4i%2B6%29%2F13 =


= %288%2Bi%29%2F13 = 8%2F13+%2B+i%2A%281%2F13%29.


The modulus of this complex number is 

sqrt%28%288%2F13%29%5E2+%2B+%281%2F13%29%5E2%29 = sqrt%2864%2F169+%2B+1%2F169%29%29 = sqrt%2865%2F169%29 = sqrt%2865%29%2F13.


Answer.  |(1+2i)/(2+3i)| = sqrt%2865%29%2F13.


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On complex numbers see the lessons in this site
    - 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


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.


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Be aware: there is an ERROR in calculations by @josgarithmetic.