How do you simplify #(18+3i)/(18-3i)#?

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Answer 1 · 2015-12-13T13:54:36

#\frac{35+12i}{37}#

First of all, you can simplify everything by a factor #3#:

#\frac{18+3i}{18-3i}=\frac{6+i}{6-i}#

Then, you can do something similar to razionalization with roots: multiply and divide by #6+i#:

#\frac{6+i}{6-i} \frac{6+i}{6+i} = \frac{(6+i)^2}{(6+i)(6-i)}#

Now use the fact that #(a+b)(a-b)=a^2-b^2#:

# \frac{(6+i)^2}{6^2 - i^2}#

Now remember that #i^2=-1#:

# \frac{(6+i)^2}{6^2 +1} =\frac{36+12i-1}{37}=\frac{35+12i}{37}#