# Homework Help: Identify the imaginary part

1. Jan 31, 2006

### UrbanXrisis

I am supposed to identify the imaginary part (marked in bold) of each expression, just wanted to see if I got them correct:

1. (1+i)+(1-i) ..........................0
2. (5+i)+(1+5i) ..........................6
3. (5+i)-(1-5i) ..........................6
4. 1+2i+3+4i+5 ..........................6
5. $$S=1+\frac{1}{2i}+\frac{1}{4}+\frac{1}{8i}+\frac{1}{16}+\frac{1}{32i}+\frac{1}{64}+...$$

using the geometric series: (.5)/(1-.25)=2/3=1/1.5i
does this mean the imaginary part is 1.5 since the i is in the denominator? or is it 1.5?

6. a and b are constants
a+bi ..........................b
7. (a+ib)^2 ..........................2ab
8. (a+bi)(b+ia) ..........................a^2+b^2
9. (ia)^3 ..........................-a^3
10. a(a+i)(a+2i) ..........................3a^2

are these correct?

Last edited: Jan 31, 2006
2. Jan 31, 2006

### VietDao29

The way I see it, everything is correct except the 5th one
If the complex number is in the form: a + ib then a is the real part, and b is the imaginary part. You must do something to make it have the form a + ib, i.e i cannot be in the denominator, it must be in the numerator.
Hint:
$$\frac{a}{ib} = \frac{ia}{i ^ 2b} = -\frac{ia}{b}$$
Multiply both numerator and denominator by i.

3. Jan 31, 2006

### UrbanXrisis

then it is -1/15?

4. Jan 31, 2006

### VietDao29

The first term is -1 / 2, and the common ratio is still 1 / 4.
It should be -2 / 3. Shouldn't it?

5. Jan 31, 2006

### HallsofIvy

The imaginary part of
$$S=1+\frac{1}{2i}+\frac{1}{4}+\frac{1}{8i}+\frac{1} {16}+\frac{1}{32i}+\frac{1}{64}+...$$
is
$$S=\frac{1}{2i}+\frac{1}{8i}+\frac{1}{32i}+...$$
which is indeed, a geometric sequence,
$$S= \frac{1}{2i}\left(1+ \frac{1}{4}+ \frac{1}{16}+...\right)$$
having sum
$$\frac{1}{2i}\frac{1}{1- \frac{1}{4}}$$
$$= \frac{1}{2i}\frac{4}{3}= \frac{2}{3}\frac{1}{i}$$
which is what you got. The only question is "what do you do with that 1/i?"

Well, i*i= -1, of course, so i(-i)= -(-1)= 1. -i is the multiplicative inverse of i: 1/i= -i.
$$\frac{2}{3}\frac{1}{i}= -\frac{2}{3}i$$.