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Mathematics
OpenStudy (anonymous):

let u=5i and z = 1+2i what is u/z

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?

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

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

OpenStudy (anonymous):

It's \[(a +bi)(a-bi) \]\[= a^2 +(ab)i - (ab)i - b^2i^2\]\[=a^2 - b^2i^2\]\[=a^2 -b^2(-1)\]\[=a^2+b^2\]

OpenStudy (anonymous):

ok thanks polpak and myininaya

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