Determine the squar roots
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pka
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Let's write as a mathematician would: \(\displaystyle 1-\sqrt3~i\).
\(\displaystyle 1-\sqrt3~i=2[\cos(-\pi/3)+i\sin(-\pi/3)]\)
So one square root is \(\displaystyle \sqrt2[\cos(-\pi/6)+i\sin(-\pi/6)]\)
How do we get the other? Hint \(\displaystyle \pi\) is needed.
Hello, poplap!
I hope I understood the question . . .
Let: .\(\displaystyle a + bi \:=\:\sqrt{1 - i\sqrt{3}}\) ..where \(\displaystyle a\) and \(\displaystyle b\) are real.
Then: .\(\displaystyle (a + bi)^2 \:=\:1 - i\sqrt{3} \quad\Rightarrow\quad (a^2-b^2) + 2abi \:=\:1-i\sqrt{3}\)
Equate real and imaginary components: .\(\displaystyle \begin{Bmatrix}a^2 - b^2 \:=\:1 & [1] \\ 2ab \:=\:\text{-}\sqrt{3} & [2] \end{Bmatrix}\)
From [2]: .\(\displaystyle b \:=\:\text{-}\dfrac{\sqrt{3}}{2a}\;\;[3]\)
Substitute into [1]: .\(\displaystyle a^2 - \left(\text{-}\dfrac{\sqrt{3}}{2a}\right)^2 \:=\:1 \quad\Rightarrow\quad a^2 - \dfrac{3}{4a^2} \:=\:1\)
. . \(\displaystyle 4a^4 - 4a^2 - 3 \:=\:0 \quad\Rightarrow\quad (2a^2 + 1)(2a^2-3) \:=\:0 \)
We have two equations to solve:
. . \(\displaystyle 2a^2 + 1 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\text{-}\frac{1}{2} \;\;\text{ but }a\text{ must be real.}\)
. . \(\displaystyle 2a^2 - 3 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\pm\dfrac{\sqrt{3}}{\sqrt{2}} \)
Substitute into [3]: .\(\displaystyle b \:=\:-\dfrac{\sqrt{3}}{2(\pm\frac{\sqrt{3}}{\sqrt{2}})} \:=\:\mp\dfrac{1}{\sqrt{2}}\)
Hence: .\(\displaystyle a + bi \;=\;\pm\dfrac{\sqrt{3}}{\sqrt{2}} + \mp\dfrac{1}{\sqrt{2}}i \)
Therefore: .\(\displaystyle \sqrt{1-i\sqrt{3}} \;=\;\pm\left(\dfrac{\sqrt{3} - i}{\sqrt{2}}\right)\)
I hope I understood the question . . .
Let: .\(\displaystyle a + bi \:=\:\sqrt{1 - i\sqrt{3}}\) ..where \(\displaystyle a\) and \(\displaystyle b\) are real.
Then: .\(\displaystyle (a + bi)^2 \:=\:1 - i\sqrt{3} \quad\Rightarrow\quad (a^2-b^2) + 2abi \:=\:1-i\sqrt{3}\)
Equate real and imaginary components: .\(\displaystyle \begin{Bmatrix}a^2 - b^2 \:=\:1 & [1] \\ 2ab \:=\:\text{-}\sqrt{3} & [2] \end{Bmatrix}\)
From [2]: .\(\displaystyle b \:=\:\text{-}\dfrac{\sqrt{3}}{2a}\;\;[3]\)
Substitute into [1]: .\(\displaystyle a^2 - \left(\text{-}\dfrac{\sqrt{3}}{2a}\right)^2 \:=\:1 \quad\Rightarrow\quad a^2 - \dfrac{3}{4a^2} \:=\:1\)
. . \(\displaystyle 4a^4 - 4a^2 - 3 \:=\:0 \quad\Rightarrow\quad (2a^2 + 1)(2a^2-3) \:=\:0 \)
We have two equations to solve:
. . \(\displaystyle 2a^2 + 1 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\text{-}\frac{1}{2} \;\;\text{ but }a\text{ must be real.}\)
. . \(\displaystyle 2a^2 - 3 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\pm\dfrac{\sqrt{3}}{\sqrt{2}} \)
Substitute into [3]: .\(\displaystyle b \:=\:-\dfrac{\sqrt{3}}{2(\pm\frac{\sqrt{3}}{\sqrt{2}})} \:=\:\mp\dfrac{1}{\sqrt{2}}\)
Hence: .\(\displaystyle a + bi \;=\;\pm\dfrac{\sqrt{3}}{\sqrt{2}} + \mp\dfrac{1}{\sqrt{2}}i \)
Therefore: .\(\displaystyle \sqrt{1-i\sqrt{3}} \;=\;\pm\left(\dfrac{\sqrt{3} - i}{\sqrt{2}}\right)\)