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a) Find the square roots of $3+4i$.

 

b) Find all cube roots of 8i.
 

where i=sqrt-1

michaelcai  Dec 12, 2017
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 #2
avatar+18829 
+2

Algrebra Help

 

a) Find the square roots of \( 3+4i\).

\(\begin{array}{rcll} z= a+bi &=& 3+4i \\ |z| &=& \sqrt{a^2+b^2} \\ &=& \sqrt{3^2+4^2} \\ &=& 5 \\ \sin(\varphi) &=& \frac{b}{|z|} \\ &=& \frac{4}{5} \\ &=& 0.8 \\ \varphi &=& \arcsin(0.8) +2k\pi \\\\ z &=& |z|e^{i\varphi} \\ \mathbf{z} & \mathbf{=} & \mathbf{5e^{i\cdot ( \arcsin(0.8)+2k\pi) }} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \sqrt{3+4i} \\ &=& (3+4i)^{\frac12} \\ &=& 5^{\frac12} \left( e^{i\cdot ( \arcsin(0.8)+2k\pi) }\right)^{\frac12} \\ &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)+2k\pi}{2}\right)} \\ &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}+k\pi\right)} \quad & | \quad k =(0,1) \\\\ \sqrt{3+4i} &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}\right)} \quad & | \quad k = 0 \\ &=& \sqrt{5} \cdot \left( \cos( \frac{\arcsin(0.8)}{2} ) + i \cdot \sin(\frac{\arcsin(0.8)}{2}) \right) \\ &=& 2.23606797750 \cdot \left( 0.89442719100 + i \cdot 0.44721359550 \right) \\ &=& 2+i\cdot 1 \\\\ \sqrt{3+4i} &=& \sqrt{5}\cdot e^{i\cdot \left( \frac{\arcsin(0.8)}{2}+\pi\right)} \quad & | \quad k = 1 \\ &=& \sqrt{5} \cdot \left( \cos( \frac{\arcsin(0.8)}{2}+\pi ) + i \cdot \sin(\frac{\arcsin(0.8)}{2}+\pi) \right) \\ &=& \sqrt{5} \cdot \left( -\cos( \frac{\arcsin(0.8)}{2}) - i \cdot \sin(\frac{\arcsin(0.8)}{2}) \right) \\ &=& 2.23606797750 \cdot \left( -0.89442719100 - i \cdot 0.44721359550 \right) \\ &=& -2-i\cdot 1 \\\\ \hline \end{array}\)

 

All 2nd roots of \(3+4 i\) :

\(2+i \\ -2-i \)

 

laugh

heureka  Dec 12, 2017
 #3
avatar+18829 
+1

Algrebra Help

 

b) Find all cube roots of 8i.

\(\begin{array}{|rcll|} \hline && \sqrt[3]{8i} \\ &=& \sqrt[3]{8}\cdot \sqrt[3]{i} \\ &=& \sqrt[3]{2^3}\cdot \sqrt[3]{i} \\ &=& 2\cdot \sqrt[3]{i} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} z= a+bi &=& i \\ |z| &=& \sqrt{0^2+1^2} \\ &=& \sqrt{1} \\ &=& 1 \\ \sin(\varphi) &=& \frac{b}{|z|} \\ &=& \frac{1}{1} \\ &=& 1 \\ \varphi &=& \arcsin(1) +2k\pi \\ \varphi &=& \frac{\pi}{2} +2k\pi \\\\ z &=& |z|e^{i\varphi} \\ & = & 1\cdot e^{i\cdot ( \frac{\pi}{2}+2k\pi) } \\ \mathbf{z} & \mathbf{=} & \mathbf{ e^{i\cdot ( \frac{\pi}{2}+2k\pi) }} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \sqrt[3]{i} \\ &=& (i)^{\frac13} \\ &=& \left( e^{i\cdot ( \frac{\pi}{2}+2k\pi)}\right)^{\frac13} \\ &=& e^{i\cdot \left( \frac{\frac{\pi}{2}+2k\pi}{3} \right)} \\ &=& e^{i\cdot \left( \frac{\pi}{6}+\frac{2}{3}\cdot k\pi \right) } \quad & | \quad k =(0,1,2) \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6} \right)} \quad & | \quad k = 0 \\ &=& \cos( \frac{\pi}{6} ) + i \cdot \sin(\frac{\pi}{6} ) \quad & | \quad \frac{\pi}{6} = 30^{\circ}\\ &=& \cos( 30^{\circ} ) + i \cdot \sin(30^{\circ}) \\ &=& \frac{\sqrt{3}}{2} + i\cdot \frac12 \\ 2\cdot \sqrt[3]{i} &=& 2\cdot \left( \frac{\sqrt{3}}{2} + i\cdot \frac12 \right) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{\sqrt{3} + i} \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6}+\frac23 \pi \right)} \quad & | \quad k = 1 \\ &=& e^{i\cdot \left( \frac{5\pi}{6} \right)} \\ &=& \cos(\frac{5\pi}{6}) + i \cdot \sin(\frac{5\pi}{6}) \quad & | \quad \frac{5\pi}{6} = 150^{\circ}\\ &=& \cos( 150^{\circ} ) + i \cdot \sin(150^{\circ}) \\ &=& -\cos( 30^{\circ} ) + i \cdot \sin(30^{\circ}) \\ &=& -\frac{\sqrt{3}}{2} + i\cdot \frac12 \\ 2\cdot \sqrt[3]{i} &=& 2\cdot \left( -\frac{\sqrt{3}}{2} + i\cdot \frac12 \right) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{-\sqrt{3} + i} \\\\ \sqrt[3]{i} &=& e^{i\cdot \left( \frac{\pi}{6}+2\cdot \frac23 \pi \right)} \quad & | \quad k = 2 \\ &=& e^{i\cdot \left( \frac{3\pi}{2} \right)} \\ &=& \cos(\frac{3\pi}{2}) + i \cdot \sin(\frac{3\pi}{2}) \quad & | \quad \frac{3\pi}{2} = 270^{\circ}\\ &=& \cos( 270^{\circ} ) + i \cdot \sin(270^{\circ}) \\ &=& 0 + i \cdot \sin(90^{\circ}) \\ &=& -i \\ 2\cdot \sqrt[3]{i} &=& 2\cdot( -i ) \\ \mathbf{2\cdot \sqrt[3]{i}} &\mathbf{=}& \mathbf{-2i} \\\\ \hline \end{array}\)


All 3rd roots of \(8 i\) :
\(\sqrt{3} + i \\ -\sqrt{3} + i \\ -2i\)

 

laugh

heureka  Dec 12, 2017
 #4
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0

Simplify the following:
sqrt(4 i + 3)

 

Express 4 i + 3 as a square using 4 i + 3 = 4 + 4 i + i^2, then look to factor.
3 + 4 i = 4 + 4 i - 1 = 4 + 4 i + i^2 = (i + 2)^2:
 sqrt((i + 2)^2 ) 

                                                                                                                                                     

For all complex z with Re(z)>0, sqrt(z^2) = z.
Cancel exponents. sqrt((2 + i)^2) = i + 2:
2 + i    and    -2 - i

Guest Dec 12, 2017
 #5
avatar
0

Simplify the following:
(8 i)^(1/3)
 
Express 8 i as a cube using 8 i = 3 sqrt(3) + 9 i + 3 sqrt(3) i^2 + i^3, then look to factor.

8 i = 3 sqrt(3) + 9 i - 3 sqrt(3) - i = 3 sqrt(3) + 9 i + 3 sqrt(3) i^2 + i^3 = (sqrt(3))^3 + 3 (sqrt(3))^2 i + 3 sqrt(3) i^2 + i^3 = (sqrt(3) + i)^3:
 ((sqrt(3) + i)^3 )^(1/3)
 

For all complex z with -π/3 Cancel exponents. ((sqrt(3) + i)^3)^(1/3) = sqrt(3) + i:

sqrt(3) + i      -sqrt(3) + i        -2i

Guest Dec 12, 2017

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