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Problem I am stuck on, simplifying radicals in denominator again

GrannySmith

New member
Jan 5, 2014
5
:mad: 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

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
-√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...

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

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
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

New member
Jan 5, 2014
5
You guess correctly that this is where you made your mistake...

[tex]\displaystyle \begin{align*} -\sqrt{11}i\cdot \sqrt{11}i &= - \left( \sqrt{11} \right) ^2 i^2 \\ &= -11 \left( -1 \right) \\ &= + 11 \end{align*}[/tex]
You're awesome! Makes so much sense now.

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
You're awesome! Makes so much sense now.

(-√6i)(-√14i) would be (√84i^2) correct? Then the i^2 would make it -√84? Just wanna make sure I fully understand this.
Yes, but also remember

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

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.