3

Simplify $(1-\sqrt{11}i)\sqrt{5-\sqrt{11}i}$

My solution:

By using half angle formula of the tangent function for the second number in the product we have $\arg(1-\sqrt{11})+\arg(\sqrt{5-\sqrt{11}i})=\arctan(-\sqrt{11})+\arctan(-\frac1{\sqrt{11}})=-\frac{\pi}{2}$ and the magnitude of the number is $$\sqrt{12}\sqrt{\sqrt{36}}=6\sqrt2.$$

So the number is $-6\sqrt2 i.$

Is my answer correct? Can you suggest a simpler way? Thanks.

Robert Shore
  • 23,332
Bob Dobbs
  • 10,988

2 Answers2

3

Since $5-\sqrt {11}i$ has $2$ square roots and we don't know which square root is meant. I assume that the square root can be both.

Here is an alternative solution:

\begin{align} \left(1-\sqrt{11}i \right)\sqrt{5-\sqrt{11}i} &=\pm \sqrt {\left(1-\sqrt{11}i \right)^2\left(5-\sqrt{11}i \right) }\\ &=\pm \sqrt{\left( -10-2\sqrt{11}i\right)\left(5-\sqrt{11}i \right) } \\ &=\pm \sqrt{-2\left( 5+\sqrt{11}i\right)\left(5-\sqrt{11}i \right) } \\ &=\pm\sqrt{-2(36)} \\ &=\pm6\sqrt2 i \end{align}

Li Kwok Keung
  • 3,744
1

$5- i \sqrt {11}$ has square roots, one of them has positive real part. I looked for it in the form $a + b i \sqrt{11}$ on the off chance that $a,b$ might be rational.

$$ \left( a + b i \sqrt{11} \right)^2 = (a^2 - 11 b^2) + 2ab i \sqrt{11} $$

so we need $a^2 - 11 b^2 = 5$ and $2ab = -1.$ Well, $4 a^2 b^2 = 1$ and $44 a^2 b^2 = 11,$ while $a^4 - 22 a^2 b^2 + 121b^4 = 25.$ Add, we see $a^4 + 22 a^2 b^2 + 121b^4 = 36$ and, with real $a,b,$ that $a^2 + 11 b^2 = 6.$ Adding and subtracting, $2 a^2 = 11$ and $22b^2 = 1.$ We are taking $a>0,$ so $b < 0.$

$a = \sqrt {\frac{11}{2}}$ while $b = - \sqrt {\frac{1}{22}}.$

Finally we get

$$ a + b i \sqrt{11} = \sqrt {\frac{11}{2}}- i \sqrt {\frac{1}{2}} $$

As a result the quantity in the question is

$$ ( 1 - i \sqrt{11} ) \left( \sqrt {\frac{11}{2}}- i \sqrt {\frac{1}{2}}\right) $$ which comes out

Will Jagy
  • 139,541