2

I have this problem :

$$x^2=i$$

The args = $\pi/2$.

$r = |z| = \sqrt{0^2+i^2}=\sqrt{i^2}=i$

for $$z_0=i((\cos (\pi/2)/2)+isin(\pi/2)/2)) = i(\frac{\sqrt{2}}{2})+i\frac{\sqrt{2}}{2})=i\frac{\sqrt{2}}{2}+i^2(\frac{\sqrt{2}}{2})$$

For some reason I don't get the result shown in the book, for some reason the book use $1$ instand of $i$ meaning that $\rightarrow$ $r=i=1$

I don't understand why, Any ideas?, Thanks.

JaVaPG
  • 2,716
  • 1
    $|z| = \sqrt{0^2 + 1^2} = 1$. (Note that $|z|$ is always positive real; measuring the distance between $z$ and $0$ in the complex plane.) – mrf Oct 10 '14 at 15:58
  • 2
    By definition, the absolute value $|z|$ of a given complex number $z=x+iy$ (for some $x,y\in \Bbb R$) is $\sqrt{x^2+y^2}$, you're setting $y=i$ when you should be setting $y=1$. See this answer for a general method to solve this sort of equation. – Git Gud Oct 10 '14 at 15:58
  • Oh, I feel so stupid, Thanks for help. – JaVaPG Oct 10 '14 at 16:03

3 Answers3

2

The modulus of $i$ is $1$: for every $z\in\mathbb{C}$, $|z|$ is a nonnegative real number: $$ |z|=\sqrt{z\bar{z}} $$ For $z=i$, $$ |i|=\sqrt{i\bar{i}}=\sqrt{i(-i)}=\sqrt{-i^2}=\sqrt{1}=1 $$ The argument is indeed $\pi/2$, so the square roots of $i$ are $$ 1\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)= \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} $$ and its opposite $$ 1\left(\cos\left(\frac{\pi}{4}+\pi)\right)+ i\sin\left(\frac{\pi}{4}+\pi)\right)\right) =-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} $$

egreg
  • 238,574
0

Note $|z| = \sqrt{ \bar{z} \cdot z} = \sqrt{a^2 + b^2}$ when $z = a+ib$ and $\bar{z} = a-ib$ is the complex conjugate. So, if $z = i$, say, then $|z| = \sqrt{ (-i)\cdot i} = \sqrt{1} = 1$ (alternative approach, $z = \sqrt{ 0^2 + 1^2} = 1$).

Tom
  • 9,978
0

let $x=a+bi$, then solve the equation $(a+bi)^2=i$ after expanding we will get $a^2-b^2+2abi=i$ or equivalent to the system
$2ab=1$
$a^2-b^2=0$

Dr. Sonnhard Graubner
  • 95,283