need help

Hello Melody,

i am not certain,

if      $z =-i$

then  $z^2 = (-i)\cdot(-i) = (-i)^2 = i^2 = -1$

if      $z^2 = -1$

then $z = \sqrt{-1}= i$, but z can also be -i

                                 this  is not equivalent (<=>)

81^(1/4)  =  $\sqrt[4]{81}$ = 3

(-81)^\frac{1}{4}

Firstly there is no real solution for this because no real number to the power of 4 is going to be negative.

But I will give you the complex solution :)

$(-81)^\frac{1}{4}\\ =(-1)^\frac{1}{4}*(81)^\frac{1}{4}\\ =(-1)^\frac{1}{4}*3\\ =((-1)^\frac{1}{2})^\frac{1}{2}*3\\ =i^\frac{1}{2}*3$

          Now

          $\sqrt{i}=\sqrt{\frac{1+2i-1}{2}}=\sqrt{\frac{1+2i+i^2}{2}}=\sqrt{\frac{(1+i)^2}{2}}=\frac{1+i}{\sqrt2}\\ $

$=3*\sqrt{i}\\ =\frac{3(1+i)}{\sqrt2}\\ =\frac{3\sqrt2 (1+i)}{2}\\$

(-81)^\frac{1}{4}

$\begin{array}{|rcll|} \hline x&=& \sqrt[4]{-81} \\ &=& (-81)^\frac{1}{4} \\ &=& (81)^\frac{1}{4} \cdot (-1)^\frac{1}{4} \\ &=& 3 \cdot (-1)^\frac{1}{4} \quad & | \quad e^{i\pi} = -1\ ! \quad (\text{Euler's Formula for Complex Numbers}) \\ x_1 &=& 3\cdot (e^{i\pi})^\frac{1}{4} \\ &=& 3\cdot e^{i\frac{\pi}{4}} \\ &=& 3\cdot [\cos(\frac{\pi}{4}) + i\cdot \sin(\frac{\pi}{4})] \quad & | \quad \cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})= \frac{\sqrt{2}}{2} \\ &=& 3\cdot (\frac{\sqrt{2}}{2}+ i\cdot \frac{\sqrt{2}}{2} ) \\ &=& \frac{3 \sqrt{2}}{2}(1+i) \\\\ x_2 &=& 3\cdot e^{i(\frac{\pi}{4}+\frac{\pi}{2} )} \\ x_2 &=& 3\cdot e^{\frac{3}{4} i \pi } \\\\ x_3 &=& 3\cdot e^{i(\frac{\pi}{4}+2\cdot\frac{\pi}{2} )} \\ x_3 &=& 3\cdot e^{\frac{5}{4} i \pi } \\\\ x_4 &=& 3\cdot e^{i(\frac{\pi}{4}+3\cdot\frac{\pi}{2} )} \\ x_4 &=& 3\cdot e^{\frac{7}{4} i \pi } \\\\ \hline \end{array}$

Hello Melody,

i am not certain,

if z =-i then z² = -1

if z² = -1 then z = $\sqrt{-1}$ = i, but z can also be -i