Imaginary numbers/complex numbers

• Aug 24th 2009, 01:53 PM
rel85
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

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

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

iii) Find
$2 / 1 - i$

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

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

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

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

Regards,
• Aug 24th 2009, 02:07 PM
pickslides
Quote:

Originally Posted by rel85

1)i) Simplify

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

$i = \sqrt{-1}$

Therefore

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

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

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

and so on

Quote:

Originally Posted by rel85

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

Group like terms

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

The rest of your problems use similar ideas.
• Aug 24th 2009, 03:01 PM
Plato
Memorize this table.
$\begin{gathered} i^{ - 1} = - i \hfill \\ i^0 = 1 \hfill \\
i^1 = i \hfill \\ i^2 = - 1 \hfill \\ i^3 = - i \hfill \\ \end{gathered}$

Once you have done that completely, then powers are very easy.

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

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

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

BTW: If you have more than one character in an exponent, set off the whole exponent in braces.
$$i^{-3}$$ gives $i^{-3}$ instead of $i^-3$.
• Aug 24th 2009, 03:06 PM
Krahl
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

$i^{73}=(i^{72})(i^1)=(i^{4.18})(i)=1.i=i$
so
$i^9=(i^8)(i)=(i^{4.2})(i)=1.i=i$
and
$i^{-9}=\frac{1}{i^9}=\frac{1}{i}=(\frac{1}{i})(\frac{i }{i})=\frac{i}{i^2}=-i$
since i^2=-1
• Aug 24th 2009, 03:10 PM
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
• Aug 24th 2009, 03:33 PM
Plato
Quote:

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: $\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

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

So $\frac{{1 + i}}
{{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}}
{25}$
• Aug 25th 2009, 10:42 AM
rel85
Quote:

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

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

So $\frac{{1 + i}}
{{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}}
{5}$

Thanks guys for the input so far.

So I have:

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

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

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

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,
• Aug 25th 2009, 11:22 AM
Plato
Quote:

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?
$\left| {a + bi} \right| = \sqrt {a^2 + b^2 }$
So $\left| {3 -4i} \right|= \sqrt {9 + 16 }=5$.

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