The first question is pretty fun, hopefully you'll agree.

Remember that any number raised to the first power is just itself; *a*^{1}=1 for all numbers *a*, so *i*^{1 }= *i*. Remember also that the number *i* is defined so that *i*^{2 }= -1; that is, *i* is the square root of negative one.

Continuing, we see that i^{3} is just *i*^{2}(*i*) = -1(*i*) = -*i *since *i*^{2}= -1. Finally, *i*^{4} = *i*^{2}(*i*^{2})= -1(-1) = 1. Now, do you also remember that any (nonzero!) number raised to the 0th power is just 1? So, *i*^{0}* = i*^{4}* =* 1.

What about *i*^{5}? It's just *i*^{4} times *i*, or in other words, 1 times *i*, so that's just *i. *Perhaps you're starting to see a pattern:

*i*^{0} = *i*^{4} = 1

*i*^{1} = *i*^{5} = *i*

*i*^{2} = ?? = -1

*i*^{3} = ?? = -*i*

What's the relationship between 0 and 4, so that *i* raised to both of these powers is the same? What's the relationship between 1 and 5 so that *i* raised to both of these powers is the same, but different from *i* raised to 0 or 4?

Well, 0 and 4 are both multiples of 4:

0 = 0(4), 4 = 1(4),

while 1 and 5 are both 1 more than a multiple of 4:

1 = 0(4) + 1, 5 = 1(4) + 1

So, back to the original question: *i*^{240 }= ??

Well, 240 = 60(4), so we can think about *i*^{240} as *i*^{4} multiplied by itself 60 times. But this is just 1 multiplied by itself 60 times, so the answer is 1.

If you can visualize the complex plane (the x-axis becomes the "real axis", while the y-axis becomes the "imaginary axis"), then multiplication by *i* simply rotates the plane 90° counterclockwise, sending 1 to *i*, *i* to -1, etc. Rotate 4 times and you're back where you started! That's why we only care about the remainder when dividing 240 by 4.

For the second question, we have to "multiply by the conjugate" to simplify this expression. The conjugate that we care about is that of the denominator, 1+*i*, whose conjugate is 1-*i *(just change the sign in front of the imaginary part).

Why in the world would we care about conjugates?

Look at multiplying the denominator by its conjugate:

(1+*i*)(1-*i*) = 1(1) + 1(-*i*) + *i*(1) + i(-i) = 1 + -*i* + *i* -*i*^{2} = 1 - *i* + *i* -(-1) = 2.

Notice that the *i* and -*i* in the middle cancel each other out, so we get a real number (that's why we used the conjugate; it's designed to do this). This will be the denominator of our answer.

So, we multiplied the denominator by the conjugate of the denominator. Anything we do to the bottom we have to do to the top. So, the numerator will be

2*i*(1-*i*) = 2*i*(1) + 2*i*(-*i*) = 2*i* - 2*i*^{2} = 2*i* - 2(-1) = 2 + 2*i = *2(1+i). The two cancels with the two in the denominator, so the final answer is 1+*i*.

Remember that if somebody tells us that 6/3 = 2, we can check their work by verifying that 3(2) = 6. We can do this here, by checking that (1+*i*)(1+*i*) = 2*i, *since we just calculated that 2*i*/(1+*i*) = 1+*i*.

Nancy L.

12/16/13