+0  
 
0
332
1
avatar

Find all complex solutions such that z^4 = -4 Note: All solutions should be expressed in the form a+bi, where a and b are real numbers.

 

I took a square root on both sides, and then got z^2 = +/- 2i (sorry, don't know how to use latex)

 

from there, I know I have to take a square root on both sides again, and I get the square root of 2, but I really don't know the way to take the square root of i, which is where I need help at. If you're kind enough to help, it would be great!

 May 26, 2022
 #1
avatar+26389 
+2

Find all complex solutions such that z^4 = -4
Note: All solutions should be expressed in the form a+bi,
where a and b are real numbers help!

 

\(\huge{z^2 = \pm 2i}\)

 

\(\begin{array}{|rcll|} \hline z^2 &=& \pm 2i \quad | \quad z = a+bi \\ \left(a+bi\right)^2 &=& \pm 2i \\ a^2+2abi+b^2i^2 &=& \pm 2i \quad | \quad i^2 = -1 \\ a^2+2abi-b^2 &=& \pm 2i \\ \mathbf{a^2-b^2+2abi} &=& \mathbf{\pm 2i} \\ \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{+2i}\\ a^2-b^2+2abi &=& 0+2i \quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& 2 \\ \mathbf{b} &=& \dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_1 &=& 1 \\ a_2 &=& -1 \\ b_1 &=& \dfrac1a_1 \\ b_1 &=& 1 \\ b_2 &=& \dfrac1a_2 \\ b_2 &=& -1 \\ \hline \end{array}\\ \hline \end{array} &\text{or}& \begin{array}{|rcll|} \hline a^2-b^2+2abi &=& \mathbf{-2i}\\ a^2-b^2+2abi &=& 0-2i\quad | \quad \text{compare}\\ \begin{array}{|rcll|} \hline 2ab&=& -2 \\ \mathbf{b} &=& -\dfrac1a \\ \hline \end{array} &\text{and}& \begin{array}{|rcll|} \hline a^2-b^2 &=& 0 \\ a^2 &=& b^2 \\ a^2 &=& \left(-\dfrac1a\right)^2 \\ a^2 &=& \left(\dfrac1a\right)^2 \\ a^4 &=& 1 \\ \mathbf{a} &=& \pm \mathbf{1} \\ a_3 &=& 1 \\ a_4 &=& -1 \\ b_3 &=& -\dfrac1a_3 \\ b_3 &=& -1 \\ b_4 &=& -\dfrac1a_4 \\ b_4 &=& 1 \\ \hline \end{array}\\ \hline \end{array} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline z&=& a_1+b_1i = 1+i\\ z&=& a_2+b_2i = -1-i\\ z&=& a_3+b_3i = 1 -i\\ z&=& a_4+b_4i = -1+i\\ \hline \end{array}\)

 

laugh

 May 26, 2022

1 Online Users