Why is $z=\sqrt{2i}$ equal to $z=1+i$?

Question

i'm trying to solve an excercice but I do not undestand why $z=\sqrt{2i}$ is equal to $z=1+i$. How can I become it?

Answer 1

$$(1+i)^2=1+2i+i^2=1+2i-1=2i$$

So $(1+i)^2=2i$, this is however not the only solutions, since $-1-i$ would also do the trick. When working with real numbers $\sqrt{a}$ is defined as the positive root. Since there is no nice way to order $\mathbb{C}$ $\sqrt{2i}$ is ambigious.

Answer 2

The notation $\sqrt{2i}$ does not make much sense here because for each non-zero complex number, there exists two distinct square roots.

In order to test whether a complex number is a square root of a given one, just compute the square.

$$(1+i)² = 1² + 2i + i² = 2i$$

Answer 3

Actually, unless you are more specific concerning the meaning of $\sqrt{\ }$, it is not true that $\sqrt{2i}$ is $1+i$. What happens is that $(1+i)^2=2i$ and therefore $1+i$ is a square root of $2i$. But $2i$ has another square root, namely $-1-i$. In fact, every complex number (except for $0$) has two and only two square roots.

Answer 4

Basically you have to solve a quadratic equation $z^2-2i =0$ and it clearly has complex roots. So let $z= a+ib$

Clearly, $(a+ib)^2= 2i$

Now, you can expand LHS and equate real & imaginary parts of LHS to real & imaginary parts of RHS.

$a^2-b^2 =0$ and
$2ab = 2$

Solving this you get $z= 1+i$ or $-1-i$

Answer 5

Given a complex number $z^2=c+di$, it's possible to find its square roots $z=\pm(a+bi)$ by use of these formulae: $$a=\pm\sqrt{\frac{c\pm|z^2|}{2}}, b=\pm\sqrt{\frac{-c\pm|z^2|}{2}}$$ If you'd like to know how I got these - ask!

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In your case, $z^2=2i\to |z^2|=2, c=0$

Thus, $a=\pm\sqrt{\pm1},b=\pm\sqrt{\pm1}$

Testing these values by squaring the resulting complex number gives:

$$z=1+i\to (1+i)^2=1^2-1^2+2i=2i \text{(works)}$$ $$z=1-i\to (1-i)^2=1^2-1^2-2i=-2i\text{(fail)}$$ $$z=-1+i\to (-1+i)^2=1^2-1^2-2i=-2i\text{(fail)}$$ $$z=-1-i\to (-1-i)^2=1^2-1^2+2i=2i\text{(works)}$$

So your square roots are $z=\pm(1+i)$

Answer 6

As I can see you are rather interested in how to find such a square root. For this you may use

• $z= |z|e^{i\phi}=|z|(\cos \phi + i \sin \phi)$
• $i= e^{i\frac{\pi}{2}}$
• $e^{i(\phi + 2\pi k)}=e^{i\phi}$ for $k \in \mathbb{Z}$

So, you get $$\sqrt{2i}=\sqrt{2}\left( e^{i(\frac{\pi}{2} + 2\pi k)} \right)^{\frac{1}{2}}= \sqrt{2}e^{i(\frac{\pi}{4} + \pi k)}= \pm\sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) = \pm (1+i)$$

Answer 7

Looking for the roots of:

$z^2 -2i =0$.

$z^2=2e^{i((π/2)+k2π))}$, $k$ an integer .

$z_k =√2e^{i((π/4) +kπ)}$, $k=0,1$.

The 2 distinct roots mod(2π) are;

$z_0= √2e^{iπ/4}$, and $z_1=√2e^{i(5/4)π}$.

$z_0= √2(\cos π/4 +i \sin π/4)=√2(1/2)(√2+i√2)=1+i$.

The other root is?