Simple numbers, just compute the actual value

Algebra Level 3

$\large {\color{#3D99F6}{A}\color{#EC7300}{=}\color{darkred}{\left(\sqrt{\sqrt{\sqrt{\sqrt{\color{#D61F06}{{-65536}}}}}}\right)}^{\color{#69047E}{40}} \\ {\color{magenta}{B}\color{#302B94}{=}\color{#456461}{\left(\sqrt{\sqrt{\sqrt{\sqrt{\color{#D61F06}{{-65536}}}}}}\right)}^{\color{#624F41}{80}}} }$

What type of numbers are $A,B$ ?

B is complex , A is a positive real A is complex , B is a positive real B is negative real , A is a positive real B is complex , A is a negative real A is complex , B is a negative real A is negative real , B is a positive real

1 solution

Nihar Mahajan
Sep 30, 2015

$\large{\sqrt{\sqrt{\sqrt{\sqrt{-65536}}}}=(-66536)^{\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}} = (-66536)^{\frac{1}{16}}}$

Thus we have:

$\large{A=(-66536)^{\frac{1}{16}\times 40} = (-66536)^{\frac{40}{16}} = (-66536)^{\frac{5}{2}}=(\sqrt{-1}\sqrt{-66536})^5 \\ = i^5\times (\sqrt{-66536})^5 = i\times(\sqrt{-66536})^5 \\ \Rightarrow A \ is \ a \ complex \ number}$

$\large{B=(-66536)^{\frac{1}{16}\times 80}= (-66536)^{\frac{80}{16}}= (-66536)^5 = -(66536)^5 \\ \Rightarrow B \ is \ a \ negative \ real}$

b is also a complex number. with imaginary part 0. hence, remove that option.

Aareyan Manzoor - 5 years, 8 months ago

Can you please write a report because I can't edit/remove the options because I am not a moderator.

Nihar Mahajan - 5 years, 8 months ago

But sqrt(-66538) is complex so both a and b are complex

Reynan Henry - 5 years, 8 months ago

Simple numbers, just compute the actual value

1 solution

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