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OpenStudy (anonymous):
\[\frac{5i}{1+2i}\times \frac{1-2i}{1-2i}\]
OpenStudy (anonymous):
denominator is
\[1+4=5\]
OpenStudy (anonymous):
numerator is
\[10+5i\]
OpenStudy (anonymous):
so you get
\[\frac{10+5i}{5}=2+i\]
OpenStudy (anonymous):
how did you get the denominator?
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OpenStudy (anonymous):
\[(a+bi)(a-bi)=a^2+b^2\]
OpenStudy (anonymous):
once you see it it is easy
OpenStudy (anonymous):
is that an identity
OpenStudy (anonymous):
If you foil it out you will see why it is true. It's just how conjugate pairs work.
OpenStudy (anonymous):
i did foil it out but i still dont get it
even though i just proved it to myself
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OpenStudy (anonymous):
Ok, well then just accept that it is true (since you proved it to yourself).
So if you want to rewrite a fraction such that it has a real denominator you just multiply top and bottom by the conjugate.
OpenStudy (anonymous):
ok. but is it really (a +bi)(a-bi) = a^2 +bi^2 or is it (a +bi)(a-bi) = a^2 -bi^2
myininaya (myininaya):
(a+bi)(a-bi)=a^2-abi+abi-b^2i^2
remember i^2=-1
so a^2-abi+abi-b^2(-1)=a^2+b^2