In these types of problem where you have a complex term (i.e. a term involving ‘i’) on the bottom of the fraction, your basic strategy is always the same. You multiply the top and bottom of the fraction by the “complex conjugate” of the denominator. If your denominator is a+bi, say, then you’ll be multiplying top and bottom by a-bi (this is what’s called the complex conjugate).
So, that’s your basic strategy. Here are the specifics for your question:
Start by multiplying top and bottom of your fraction by (2+i). That gives:
Multiplying out the bit in the square brackets (i.e. the denominator) gives:
Remember that i^2 = -1, so that’s:
So your answer is 2+i
The trick is to multiply the denominator(also the numerator by the conjugate of the denominator. The conjugate of a complex number, a + bi, is a – bi.
In your example, the conjugage of 2 – i, is 2 + i
Doing that, you get:
[5*(2+i)]/[(2-i)(2+i)], the numerator is 10 + 5i, while the denominator is (2-i)(2+i)= 4 + 2i -2i -i^2, but we know that i^2=-1, so (2-i)(2+i)=4 + 0 + 1=5
So the new expression in the form of a + bi is
(10 + 5i)/5= 2 + i
Multiply the numerator and denominator by the conjugate of 2 – i
5/(2 – 1)
[5(2 + i)]/[(2 – i)(2 + i)]
[5(2 + i)]/(4 + 2i – 2i – 1^2)
[5(2 + i)]/(4 + 1)
2 + i
In your original equation, you wrote 2x where it is supposed to be 3x but it came out right anyway. Okay, think of it this way. -51 is the same as (-1)*51 right? So inside that root you have √(-1)51 and if you know how to work with roots, you know that means the same as √(-1)*√51. Well, √-1 is i right? So √-51 is the same has i*√51. Which means you have (-3 +/- i*√51) all over 6. This simplifies to (-1/2) +/- (√51 / 6)*i. Kinda nasty one…
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Express 5/(2-i) in simplest a+bi form?
5/(2 — i) = (2+i)(2–i) / (2–i) = 2 + i