SOLUTION: 4-2i ÷ 3+i -5-3i ÷ 2-2i

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 4-2i ÷ 3+i -5-3i ÷ 2-2i      Log On


   

Question 179579: 4-2i ÷ 3+i


-5-3i ÷ 2-2i
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
%284-2i%29%283-i%29+%2F+%283%2Bi%29%283-i%29
:
%2812-4i-6i%2B2i%5E2%29%2F%289%2B1%29
:
%2812-10i-2%29%2F10
:
%2810-10i%29%2F10
:
highlight%281-i%29
:
:
:
%28-5-3i%29%282%2B2i%29+%2F+%282-2i%29%282%2B2i%29
:
%28-10-10i-6i-6i%5E2%29%2F%284%2B4%29
:
%28-4-16i%29%2F8
:
highlight%28%28-1%2F2%29-2i%29
I will do one. and get you started on the other.. you must multiply by the conjugate to get rid of the complex number in the denominator
:
1.3+i√5/3-i√5
:so we have %281.3%2Bi%2Asqrt%285%29%29%2F%283-i%2Asqrt%285%29%29
:
so the conjugate of 3-+sqrt+%285%29 is 3+%2Bi%2Asqrt+%285%29
:

:
%283.9%2B2.3i%2Asqrt%285%29-5%29%2F%289%2B5%29
:
%281.1-2.3i%2Asqrt%285%29%29%2F14
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:
:
2.4-i√2/i√2 the conjugate for this is -i%2Asqrt%282%29 repeat what took place in the problem I just solved...multiplying both top and bottom by the conjugate