# Thread: "i" is the square root of -1

1. ## "i" is the square root of -1

Hello I am currently trying to study for this Math assessment test in a week and I ran into an issue with two of the problems, but both have that same rule given in the title.... When given the Sample Questions from the school they also gave me the answer too... So Yes I do have the answer but I do not understand how they got it..PLEASE HELP

( And Note "i" is the square root of -1 for both questions)

1. Multiply: (3-5i)(6+2i) Answer: 18 + 6i - 30i + 10 = 28 - 24i.

2. 2-3i OVER 4+3i answer: -1/25 - 18/25i

PLEASE EXPLAIN THIS FOR ME THANK YOU

2. ## Re: "i" is the square root of -1

1. Use FOIL to expand the brackets and use the fact that i^2 = -1 to simplify.

2. Multiply top and bottom by the bottom's conjugate, 4 - 3i.

3. ## Re: "i" is the square root of -1

Originally Posted by antoniolocke
Hello I am currently trying to study for this Math assessment test in a week and I ran into an issue with two of the problems, but both have that same rule given in the title.... When given the Sample Questions from the school they also gave me the answer too... So Yes I do have the answer but I do not understand how they got it..PLEASE HELP

( And Note "i" is the square root of -1 for both questions)

1. Multiply: (3-5i)(6+2i) Answer: 18 + 6i - 30i + 10 = 28 - 24i.

2. 2-3i OVER 4+3i answer: -1/25 - 18/25i

PLEASE EXPLAIN THIS FOR ME THANK YOU

$\displaystyle i^2=-1$

$\displaystyle (3-5i)(6+2i)=3\cdot6+3\cdot 2i-5i\cdot6-5i\cdot 2i=18+6i-30i-10i^2=18+6i-30i-10(-1)=18+6i-30i+10=(18+10)+i(6-30)=28-i24$

$\displaystyle \frac{2-3i}{4+3i}=\frac{(4-3i)(2-3i)}{(4+3i)(4-3i)}=\frac{8-12i-6i-9}{16+9}=\frac{-1-18i}{25}=-\frac{1}{25}-i\frac{18}{25}$

4. ## Re: "i" is the square root of -1

Originally Posted by antoniolocke
1. Multiply: (3-5i)(6+2i) Answer: 18 + 6i - 30i + 10 = 28 - 24i.
2. 2-3i OVER 4+3i answer: -1/25 - 18/25i
I will give you some shortcuts.
$\displaystyle z\cdot\overline{z}=|z|^2$.
So if $\displaystyle z=a+b\mathf{i}$ then $\displaystyle z\cdot\overline{z}=a^2+b^2$.

Here is a use for that. $\displaystyle \frac{1}{z}=\frac{\overline{z}}{|z|^2}$.
So $\displaystyle \frac{1}{4+3\mathf{i}}=\frac{4-3\mathf{i}}{25}$.

Again this is a basic idea: $\displaystyle (a+b)(x+y)=ax+ay+bx+ay$. Learn that.
Thus $\displaystyle (2-3\mathf{i})(4-3\mathf{i})=(2)(4)+(2)( -3\mathf{i})+(-3\mathf{i})(4)+(-3\mathf{i})(-3\mathf{i})=~?$