# i evaluations

• Aug 25th 2009, 06:47 PM
rel85
i evaluations
Hi,

I'd be very appreciative if someone could explain the steps and methodology to 'evaluate' the following:

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

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

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

Many regards,
• Aug 25th 2009, 07:17 PM
artvandalay11
Quote:

Originally Posted by rel85
Hi,

I'd be very appreciative if someone could explain the steps and methodology to 'evaluate' the following:

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

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

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

Many regards,

i) $(2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i)$ we just foiled
$=4-6i+6i-9i^2=4-9(-1)=4+9=13$ and since $i=\sqrt{-1}, i^2=-1$

Should there be parenthesis for the fraction in ii)? otherwise just combine the i terms

iii) $i^5(1 + i^5)$
$i^5=i^4*i=i^2*i^2*i=(-1)(-1)i=i$ so we now have

$i(1+i)=i+i^2=i-1=-1+i$

In general, to evaluate $i^n$ divide n by 4, and take the remainder as the power, for instance $i^{27}$ $\frac{27}{4}=6$ remainder 3, so $i^{27}=i^3=-i$

$i^0=1$
$i^1=i$
$i^2=-1$
$i^3=-i$
$i^4=1$ this is why we divide by 4, the cycle repeats as you see here
• Aug 25th 2009, 07:40 PM
rel85
Quote:

Originally Posted by artvandalay11
i) $(2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i)$ we just foiled
$=4-6i+6i-9i^2=4-9(-1)=4+9=13$ and since $i=\sqrt{-1}, i^2=-1$

Should there be parenthesis for the fraction in ii)? otherwise just combine the i terms

iii) $i^5(1 + i^5)$
$i^5=i^4*i=i^2*i^2*i=(-1)(-1)i=i$ so we now have

$i(1+i)=i+i^2=i-1=-1+i$

In general, to evaluate $i^n$ divide n by 4, and take the remainder as the power, for instance $i^{27}$ $\frac{27}{4}=6$ remainder 3, so $i^{27}=i^3=-i$

$i^0=1$
$i^1=i$
$i^2=-1$
$i^3=-i$
$i^4=1$ this is why we divide by 4, the cycle repeats as you see here

Am I doing something wrong for the first one?
I get $4 - 9i^2 + 6^i - 6^i$ from expanding the brackets which becomes $4 - 9i^2 = 4 - 9(-1) == 13$

Does it stop there, where is the i^2 = -1 coming from at this point?

No parenthesises for the second, just a fraction... Group like terms how?

Thanks!
• Aug 25th 2009, 07:46 PM
Krahl
$

7 - 3i / 2 + 4i
$

$7-\frac{3}{2}i+4i$

$7+(4-\frac{3}{2})i$

group the i terms as you would x terms
• Aug 25th 2009, 07:46 PM
artvandalay11
$(\sqrt{x})^2=x$

$i=\sqrt{-1}$ so square both sides

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

$7-\frac{3i}{2}+4i=7-\frac{3i}{2}+\frac{8i}{2}$

$=7+\frac{5i}{2}$