+0  
 
0
32
2
avatar

Find all complex numbers \(z\)  such that \(z^4 = -4. \)

Note: All solutions should be expressed in the form , where  and  are real numbers.

 May 2, 2021
 #1
avatar
0

We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

 

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

 

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

 May 2, 2021
 #2
avatar+25739 
+1

Find all complex numbers \(z\) such that \(z^4 = -4\).

 

\(\begin{array}{|rcll|} \hline z^4 &=& -4 \quad | \quad \text{sqrt both sides} \\ z^2 &=& \pm \sqrt{-4} \\ z^2 &=& \pm \sqrt{(-1)*4} \\ z^2 &=& \pm \sqrt{-1}*\sqrt{4} \quad | \quad \sqrt{-1} = i \\ z^2 &=& \pm 2i \quad | \quad \text{sqrt both sides} \\ \mathbf{z} &=& \mathbf{\pm \sqrt{\pm 2i}} \\ &&\boxed{z_1 = \sqrt{2i}\\z_2 = \sqrt{-2i}\\ z_3 = -\sqrt{2i}=-z_1\\z_4 = -\sqrt{-2i}=-z_2 } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (1+i)^2 &=& 1+2i+i^2 \quad | \quad \mathbf{i^2 =-1} \\ (1+i)^2 &=& 1+2i-1 \\ \mathbf{(1+i)^2} &=& \mathbf{2i} \\ \hline \end{array} \begin{array}{|rcll|} \hline (1-i)^2 &=& 1-2i+i^2 \quad | \quad \mathbf{i^2 =-1} \\ (1-i)^2 &=& 1-2i-1 \\ \mathbf{(1-i)^2} &=& \mathbf{-2i} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline z_1 &=& \sqrt{2i} \quad | \quad \mathbf{2i = (1+i)^2} \\ z_1 &=& \sqrt{(1+i)^2}\\ \mathbf{z_1} &=& \mathbf{1+i} \\\\ z_2 &=& \sqrt{-2i} \quad | \quad \mathbf{-2i = (1-i)^2} \\ z_2 &=& \sqrt{(1-i)^2}\\ \mathbf{z_2} &=& \mathbf{1-i} \\\\ z_3 &=& -z_1 \\ z_3 &=& -(1+i) \\ \mathbf{z_3} &=& \mathbf{-1-i} \\\\ z_4 &=& -z_2 \\ z_4 &=& -(1-i) \\ \mathbf{z_4} &=& \mathbf{-1+i} \\ \hline \end{array}\)

 

laugh

 May 3, 2021

93 Online Users

avatar
avatar
avatar
avatar
avatar