Can you help me understand how to only divide by i? Example: 14+i/i

2 Answers
KillerBunny
Jun 6, 2018

Dividing by #i# is the same as multiplying by #-i#.

Explanation:

Assuming you mean #\frac{14+i}{i}#, otherwise #14+i/i = 14+1=15#

But anyway, since you ask it, let's see how to divide by #i# in general: you can perform some sort of rationalization, multiplying and dividing by #i#:

# \frac{14+i}{i}*\frac{i}{i} = \frac{i(14+i)}{i^2}= \frac{i(14+i)}{-1} = -i(14+i)#

So, dividing by #i# is the same as multiplying by #-i#.

This makes particularly sense if you notice the periodicity of the powers of #i#:

#i^0 = 1#
#i^1 = i#
#i^2 = -1#
#i^3 = -i#
#i^4 = 1#
#i^5 = i#
#...#

In general, you have #i^{4n+k} = i^k#. But this is true for negative exponents as well!

#i^{-4} = 1#
#i^{-3} = i#
#i^{-2} = -1#
#i^{-1} = -i#
#i^0 = i#
#...#

In fact, you can write

#1/i = i^-1 = i^{4*(-1)+3} = i^3 = -i#

And this is why dividing by #i# is like multiplying by #-i#

Douglas K.
Jun 6, 2018

You multiply by #1# in the from #i/i#; this makes the divisor become #-1# because #i xx i = -1#

Explanation:

Given: #(14+i)/i#

Multiply by #1# in the form of #i/i#

#(14+i)/i i/i#

The denominator becomes #-1#:

#(i(14+i))/-1#

Use the distributive property to multiply each term in the denominator by #i#

#(14i+i^2)/-1#

We know that #i^2 = -1#:

#(14i-1)/-1#

Divide by -1:

#1 -14i#

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