Very simple question with an answer that I cannot understand:
I have $\sqrt{8i}$, which, I suppose, is the same as $\sqrt{\sqrt{-64}}$. How come that $2+2i$ is the same as $\sqrt{8i}$?
My understanding is that $\sqrt{8i}$ is the same as: (a) $\sqrt{2^3i}$ OR (b) $2\sqrt{2i}$
I'm pretty sure (a) is correct and (b) might also be correct, but how can you get from there to $2+2i$?
Thanks in advance
$\endgroup$ 38 Answers
$\begingroup$one way to see it is there exists some $z$ such that $z^2=8i$, now sub $z=a+ib$ to get $a^2-b^2+2iab=8i$, equate real and the imaginary parts, solve the simultaneous equation. if you know about the exponential form of complex numbers, this problem can be solved even quicker but I suspect you haven't learnt that yet. By the way be careful in the above statement when you say $\sqrt{\sqrt{}}$, personally its not evidently clear your allowing both the $\sqrt{}$ values to be accounted as it seems your accounting for the positive only, remember the $\sqrt{}$ of negatives don't exist in the reals but does in complex numbers.
$\endgroup$ 2 $\begingroup$HINT: $(1+i)^2=1+2i+i^2=2i\qquad\qquad$
$\endgroup$ $\begingroup$The absolute value of $8i$ is $8$, so the absolute value of its square root is $\sqrt{8} = 2\sqrt{2}$.
The angle from the real axis to the imaginary axis (where $8i$ is located) is $90^\circ$, so the angle from the real axis to the ray on which $\sqrt{8i}$ is located is half of that: $45^\circ$. Consequently it's a number of the form $a+ai$, where $a>0$. The absolute value of $a+ai$ is $\sqrt{a^2+a^2} = a\sqrt{2}$.
So we have $a\sqrt{2} = 2\sqrt{2}$, so $a=2$.
$\endgroup$ $\begingroup$You can go this way
$$ \sqrt{8} \sqrt{i}= \sqrt{8} e^{\frac{1}{2}\ln i}= \sqrt{8}\,e^{\frac{1}{2}\left(\ln|i|+i\left(\frac{\pi}{2}+2k\pi\right)\right)}= \sqrt{8}\,e^{\left(i(\frac{\pi}{4})+k\pi\right)},\quad k=0,1.$$
Note that, if you take $k=2,3,\dots$, then you go back to the same roots.
$\endgroup$ 7 $\begingroup$let $$x+iy=\sqrt{8i}$$ $$(x+iy)^2={8i}$$ $$x^2-y^2+2xyi=0+8i$$ compare real and imaginary parts of both sides: $x^2-y^2=0\;; 2xy=8$ $$x^2-y^2=0$$ square both sides $$\implies (x^2-y^2)^2 =0\implies (x^2+y^2)^2-(2xy)^2=0$$ $$\implies (x^2+y^2)^2=64 \implies (x^2+y^2)=\pm 8 $$ since sum of square of two real no. can't be negative so $$x^2+y^2=8\text{ and }x^2-y^2=0$$ solving these two eqn will give $x=y=\pm 2 $
so $$\sqrt {8i}=\pm 2(1+i)\implies 2+2i,-2-2i$$
$\endgroup$ 6 $\begingroup$$(a+bi)^2=(a+bi)(a+bi)=a^2-b^2+2abi\implies \sqrt{a^2-b^2+2abi}=a+bi$
and for conjugate $(a-bi)^2=(a-bi)(a-bi)=a^2+b^2-2abi\implies \sqrt{a^2+b^2-2abi}=a-bi$
Now if we find $a$ and $b$ satisfy appropriate equotation:
For $(a+bi)^2$:
$a^2-b^2+2abi=8i$. This is valid for any $a=b$ where $2ab=8$, or
$2a^2=8 \implies a=\pm\sqrt{4}=\pm{2} = b$. Here exists two numbers:
First $$\sqrt{8i}=2+2i$$ and second $$\sqrt{8i}=-2-2i$$
For $(a-bi)^2$:
$a^2+b^2-2abi=8i$. Because $a^2 \ge 0$ and $b^2 \ge 0$, there does not exists solution.
Hint :
$\endgroup$ $\begingroup$$$ \sqrt i = \frac1{\sqrt2}(1+i) $$
First note that $(2+2i)^2=4-4+8i=8i$. Now to be exact $$\sqrt{8i}=2\sqrt{2}e^{i(2k+1)\pi/4}\tag{*}$$ where $k\in \{0,3\}$. So actually $\sqrt{8i}$ is multi-valued just as $\sqrt{8}=\pm 2\sqrt{2}$ is multi-valued.. The identity $(*)$ comes from a famous formula proved by Euler. You can learn more about it here.
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