Question

Solve for z in z^2-i = 0?

Answer 1

I got to z = sqrt(i), at which point I substituted for sqrt(i)=a + bi. Solving gave me \[\frac{ \sqrt(2) }{ 4 } + \frac{ 1 }{ \sqrt{2} }i\]

Answer 2

But the book says the answer is

Answer 3

\[\frac{ 1 }{ \sqrt{2} }-\frac{ 1 }{ \sqrt{2} }i\]

Answer 4

No wait, make that a plus

Answer 5

\[i=\sqrt{-1}\]

Answer 6

that is all I can help you with.. I haven't taken Calculus yet...

Answer 7

don't answer questions you can't answer...

Answer 8

sorry.. I didn't see it was labeled under calculus.. Any question posted in the sub categories of math jsut show up in math...

Answer 9

\[\begin{align*}
z^2-i&=0\\\\
z^2&=i\\\\
z&=i^{1/2}\\\\
z&=\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{1/2}\\\\
z&=\begin{cases}
\cos\dfrac{\dfrac{\pi}{2}}{2}+i\sin\dfrac{\dfrac{\pi}{2}}{2}\\\\
\cos\left(\dfrac{\dfrac{\pi}{2}+\pi}{2}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+\pi}{2}\right)
\end{cases}
\end{align*}\]

Answer 10

Hold on, that second root doesn't seem to be right... It's giving me \(-i\) when I square it.

Answer 11

@SithsAndGiggles the second should be pi/4 + pi

Answer 12

5pi/4

Answer 13

Ah yes, the numerator in the argument should be \(\dfrac{\pi}{2}+2\pi\)... Thanks

Answer 14

Right, I forgot I could convert to polar form... thanks

Answer 15

yw