1. ## Imaginary numbers/complex numbers

Hi there,

Could somebody help me with these questions, explaining the solutions and maybe providing some good references for learning?

1)i) Simplify

$\displaystyle i^7, i^-3, i^9$

ii) Find
$\displaystyle (3 + 5i) + (7 - i)$

iii) Find
$\displaystyle 2 / 1 - i$

2)i) Evaluate
$\displaystyle (2 + 3i) + (2 - 3i)$

ii)
$\displaystyle 7 - 3i/2 + 4i$

iii)
$\displaystyle i^5(1 + i^5)$

I'd be very appreciative with some help in understanding/solving these types of question.

Regards,

2. Originally Posted by rel85

1)i) Simplify

$\displaystyle i^7, i^{-3}, i^9$

$\displaystyle i = \sqrt{-1}$

Therefore

$\displaystyle i \times i = i^2= \sqrt{-1} \times \sqrt{-1} = -1$

$\displaystyle i \times i \times i= i^3= \sqrt{-1} \times \sqrt{-1} \times \sqrt{-1}= -\sqrt{-1} = -i$

$\displaystyle i \times i \times i \times i = i^4= (i^2)^2 = (-1)^{2} = 1$

and so on

Originally Posted by rel85

ii) Find
$\displaystyle (3 + 5i) + (7 - i)$
Group like terms

$\displaystyle 3+7 +5i-i = 10+4i$

The rest of your problems use similar ideas.

3. Memorize this table.
$\displaystyle \begin{gathered} i^{ - 1} = - i \hfill \\ i^0 = 1 \hfill \\ i^1 = i \hfill \\ i^2 = - 1 \hfill \\ i^3 = - i \hfill \\ \end{gathered}$
Once you have done that completely, then powers are very easy.

If $\displaystyle n\ge 0$ then consider $\displaystyle i^n$.
Divide $\displaystyle n$ by 4 and take the remainder $\displaystyle j$ then $\displaystyle i^n = i^j$. You have memorized that value.
Example: $\displaystyle i^{35}=i^3=-i$ because the remainder of 35 divided by 4 is 3.

Now what if $\displaystyle n<0$ such as $\displaystyle i^{-19}$.
Change that as follows: $\displaystyle i^{-19}=\left(i^{-1}\right)^{19}=\left(-i\right)^{19}=-(-i)=i$.

Another example: $\displaystyle i^{-10}=(-i)^{10}=i^2=-1$

BTW: If you have more than one character in an exponent, set off the whole exponent in braces.
$$i^{-3}$$ gives $\displaystyle i^{-3}$ instead of $\displaystyle i^-3$.

4. i)

What you should remember about powers of i is that i to the powr of a multiple of 4 is 1. so
for example

$\displaystyle i^{73}=(i^{72})(i^1)=(i^{4.18})(i)=1.i=i$
so
$\displaystyle i^9=(i^8)(i)=(i^{4.2})(i)=1.i=i$
and
$\displaystyle i^{-9}=\frac{1}{i^9}=\frac{1}{i}=(\frac{1}{i})(\frac{i }{i})=\frac{i}{i^2}=-i$
since i^2=-1

5. iii)
multiply the top and bottom of the fraction by 1+i or -1-i, then simplify

and do the same with 2 ii) and see what you get

6. Originally Posted by Krahl
iii)
multiply the top and bottom of the fraction by 1+i or -1-i, then simplify
and do the same with 2 ii) and see what you get
It is a lot easier than that.
Just learn on simple fact: $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

Example: $\displaystyle \frac{1}{{3 - 4i}} = \frac{{3 + 4i}}{25}$

So $\displaystyle \frac{{1 + i}} {{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}} {25}$

7. Originally Posted by Plato
It is a lot easier than that.
Just learn on simple fact: $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

Example: $\displaystyle \frac{1}{{3 - 4i}} = \frac{{3 + 4i}}{5}$

So $\displaystyle \frac{{1 + i}} {{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}} {5}$
Thanks guys for the input so far.

So I have:

1. $\displaystyle i^7 = 7/4 = 1.75 - 1 x 4 = -i$

$\displaystyle i^{-3} = (i^{-1})^3 = (-i)^3 = -(-i) = i$

$\displaystyle i^9 = 9/4 = 2.25 - 2 = 0.25 x 4 = i^1 = i$

I understand ii), but I the fraction puts me off, where does the 5 come from Plato? I feel I need further instruction on this and also does Q2 require the same methods?

Regards,

8. Originally Posted by rel85
I understand ii), but I the fraction puts me off, where does the 5 come from Plato?
Do you understand the absolute value of a complex number?
$\displaystyle \left| {a + bi} \right| = \sqrt {a^2 + b^2 }$
So $\displaystyle \left| {3 -4i} \right|= \sqrt {9 + 16 }=5$.

Now take the question $\displaystyle \frac{2}{1-i}$
Because $\displaystyle \left| {1-i} \right|=\sqrt{2}$ then $\displaystyle \frac{2}{1-i}=\frac{2(1+i)}{2}=1+i$.