MHB Problem I am stuck on, simplifying radicals in denominator again

[GrannySmith]

I really hate these problems.

(2+√7)/(3-√-11). What the heck?

I start out by multiplying both sides with the conjugate again. This is where I am stuck lol. Can someone tell me what I am doing wrong while multiplying the conjugate?

(3 - √-11) is the same as 3 - √11i correct? So I multiply (3 - √11i) with (3 + √11i). I don't understand how this goes, but I tried.

3 times 3 is 9.

3 times √11i is 3√11i

-√11i times 3 is -3√11i canceling out 3√11i

-√11i times √11i is -11i? A bit confused on how this works and I'm guessing this is where I made a mistake?

On top we have (2+√7)(3 + √11i).

2 times 3 is 6

2 times √11i is 2√11i

√7 times 3 is 3√7

What would √7 times √11i be? √77i? Can you multiply a normal square root with an imaginary square root?

 

[Prove It]

-√11i times √11i is -11i? A bit confused on how this works and I'm guessing this is where I made a mistake?

You guess correctly that this is where you made your mistake...

\displaystyle \begin{align*} -\sqrt{11}i\cdot \sqrt{11}i &= - \left( \sqrt{11} \right) ^2 i^2 \\ &= -11 \left( -1 \right) \\ &= + 11 \end{align*}

 

[Jameson]

Good catch, Prove It. I delete my post with incorrect results. This still follows the pattern of $(a-b)(a+b)=(a+b)(a-b)=a^2-b^2$. In this case, $b=i \cdot \sqrt{11}$ thus $-b^2=-i^2(\sqrt{11})^2$, just like you wrote.

 

[GrannySmith]

You guess correctly that this is where you made your mistake...

\displaystyle \begin{align*} -\sqrt{11}i\cdot \sqrt{11}i &= - \left( \sqrt{11} \right) ^2 i^2 \\ &= -11 \left( -1 \right) \\ &= + 11 \end{align*}

You're awesome! Makes so much sense now.

(-√6i)(-√14i) would be (√84i^2) correct? Then the i^2 would make it -√84? Just want to make sure I fully understand this.

 

[Prove It]

You're awesome! Makes so much sense now.

(-√6i)(-√14i) would be (√84i^2) correct? Then the i^2 would make it -√84? Just want to make sure I fully understand this.

Yes, but also remember

\displaystyle \begin{align*} \sqrt{84} &= \sqrt{ 4 \cdot 21 } \\ &= \sqrt{4} \cdot \sqrt{21} \\ &= 2\sqrt{21} \end{align*}

 

[Deveno]

Just a quick note:

With two complex numbers $a+bi$ and $a-bi$ their product is:

$(a+bi)(a-bi) = a^2 - a(bi) + (bi)a - (bi)^2$.

Now the terms $-a(bi)$ and $(bi)a = a(bi)$ cancel, leaving us with:

$a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2$.

In other words, a SUM of two (real) squares can be viewed as a DIFFERENCE of two complex squares:

$a^2 + b^2 = a^2 - (bi)^2$

This unusual "trick" is what throws most people off, because we are used to thinking of squares as "always positive" (which is true for real numbers, but NOT for complex numbers, where "positive" doesn't really MEAN anything).

In general:

$\dfrac{1}{a+bi} = \dfrac{a-bi}{a^2+b^2}$

this formula is well worth remembering, since it's a real time-saver:

In your example, to compute:

$\dfrac{1}{3 - \sqrt{-11}} = \dfrac{1}{3 - \sqrt{11}i}$

so we can multiply instead of divide (because let's face it, division is a pain in the a-erm, anterior region)

instead of going through the routine of multiplying by the conjugate top and bottom, we do THIS:

$3^2 = 9$ and $(\sqrt{11})^2 = 11$ and $9 + 11 = 20$

so we have:

$\frac{1}{11}(3 + \sqrt{11}i)$

As for your "multiplication" question, it is true that:

$\sqrt{7}(\sqrt{11}i) = (\sqrt{7}\sqrt{11})i = \sqrt{77}i$

because the complex numbers form a FIELD, and this means that multiplication is associative.