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Find all complex numbers  such that  z^2=2i

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (Don't include quotes in your answer.)

 

Guest Aug 16, 2018

Best Answer 

 #1
avatar+20153 
+2

Find all complex numbers  such that  z^2=2i

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer

 

\(\begin{array}{|lrcll|} \hline & (a+bi)^2 &=& a^2 + 2abi+b^2i^2 \quad | \quad i^2 = -1 \\ & (a+bi)^2 &=& a^2 + 2abi -b^2 \\ & (a+bi)^2 &=& \underbrace{a^2-b^2}_{=0} + \underbrace{2abi}_{=2i} \quad | \quad \text{compare with } 2i \\\\ 1. & a^2-b^2 &=& 0 \\ & a^2 &=& b^2 \\ & \mathbf{a} & \mathbf{=}& \mathbf{b} \\\\ 2. & 2abi &=& 2i \quad | \quad : 2i \\ &\mathbf{ ab } & \mathbf{=}& \mathbf{1 } \quad | \quad a=b \\ &a^2 & = & 1 \\ &a & = & \pm 1 \\ \\ &b &=& a \\ & b &=& \pm 1 \\\\ & && & \mathbf{ab = 1\ ?} \\ & a &=& 1 \quad \text{ and } \quad b = 1 & ab = 1\cdot 1 = 1\ \checkmark \\ \text{or}& a &=& -1 \quad \text{ and } \quad b = -1 & ab = (-1)\cdot (-1) = 1\ \checkmark \\ \text{else}& a &=& -1 \quad \text{ and } \quad b = 1 & ab = (-1)\cdot 1 = -1\ \text{ no solution} \\ & a &=& 1 \quad \text{ and } \quad b = -1 & ab = 1\cdot (-1) = -1\ \text{ no solution} \\ \hline \end{array}\)

 

The answer is \(\mathbf{"1+i, -1-i"}\)

 

laugh

heureka  Aug 16, 2018
edited by heureka  Aug 16, 2018
 #1
avatar+20153 
+2
Best Answer

Find all complex numbers  such that  z^2=2i

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer

 

\(\begin{array}{|lrcll|} \hline & (a+bi)^2 &=& a^2 + 2abi+b^2i^2 \quad | \quad i^2 = -1 \\ & (a+bi)^2 &=& a^2 + 2abi -b^2 \\ & (a+bi)^2 &=& \underbrace{a^2-b^2}_{=0} + \underbrace{2abi}_{=2i} \quad | \quad \text{compare with } 2i \\\\ 1. & a^2-b^2 &=& 0 \\ & a^2 &=& b^2 \\ & \mathbf{a} & \mathbf{=}& \mathbf{b} \\\\ 2. & 2abi &=& 2i \quad | \quad : 2i \\ &\mathbf{ ab } & \mathbf{=}& \mathbf{1 } \quad | \quad a=b \\ &a^2 & = & 1 \\ &a & = & \pm 1 \\ \\ &b &=& a \\ & b &=& \pm 1 \\\\ & && & \mathbf{ab = 1\ ?} \\ & a &=& 1 \quad \text{ and } \quad b = 1 & ab = 1\cdot 1 = 1\ \checkmark \\ \text{or}& a &=& -1 \quad \text{ and } \quad b = -1 & ab = (-1)\cdot (-1) = 1\ \checkmark \\ \text{else}& a &=& -1 \quad \text{ and } \quad b = 1 & ab = (-1)\cdot 1 = -1\ \text{ no solution} \\ & a &=& 1 \quad \text{ and } \quad b = -1 & ab = 1\cdot (-1) = -1\ \text{ no solution} \\ \hline \end{array}\)

 

The answer is \(\mathbf{"1+i, -1-i"}\)

 

laugh

heureka  Aug 16, 2018
edited by heureka  Aug 16, 2018

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