$$(1-i)^6 \cdot \{(\sqrt3 + i)^3\}$$ to be written in the $a+bi$ form this is a linear algebra problem. I need to find the $a+bi$ format of the question.
Complex number in binomic form
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$\begingroup$
linear-algebra
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0hold up the square root is just on 3 – 2017-02-14
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0OK. What have you tried? Many ways to solve this. – 2017-02-14
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0Yeah I was trying to calculate the square root of $3+i$ ... miles easier like that ! – 2017-02-14
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0The obvious thing here is just to expand everything and collect the real and imaginary parts. Where in that process are you getting stuck? – 2017-02-14
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0so i got (8e^-3pi/2)(8e^pi/2) how do i convert to a+bi form from here? – 2017-02-14
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0@robertlikessaif why did you convert to exponential/polar form???? You csn just expand the powers using binomial theorem. No need for exponential at all. – 2017-02-14
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0$(1-i)/\sqrt{2}$ is an eigth root of unity & $(\sqrt{3}+i)/2$ is a twelth root of unity ... should be easy from there ? – 2017-02-14
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0Just expand: $-64$ – 2017-02-14
4 Answers
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$$a = (1-i)^6 = \left(\sqrt{2}e^{-i\pi/4}\right)^6 = 8e^{-i3\pi/2} = 8i$$
$$b = (\sqrt{3}+i)^3 = \left(2e^{i\pi/6}\right)^3 = 8e^{i\pi/2} = 8i$$
$$ab = -64$$
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0Best answer as this is shows exactly the reason for swapping to exponential form. – 2017-02-14
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$(1+i)^6=(-2i)^3=8i$.
$(\sqrt3+i)^3=8(\frac{\sqrt3}{2}+\frac{1}{2}i)^3=8i$
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$$(1-i)^6=1-6i+15i^2-20i^3+15i^4-6i^5+i^6=1-6i-15+20i+15-6i-1=8i$$ $$(\sqrt{3}+i)^3=\sqrt{3}^3+3\sqrt{3}^2i+3\sqrt{3}i^2+i^3=3\sqrt{3}+9i-3\sqrt{3}-i=8i$$ so expression is $-64$.
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Question: Simplify $$(1-i)^6\cdot\sqrt{(3+i)^3}\tag1$$
Let's start with $(1-i)^6$ first. Observe that $(1-i)^2=-2i$. Therefore,$$(1-i)^6=\{\color{blue}{(1-i)^2}\}^3=(\color{blue}{-2i})^3=8i\tag2$$
Now, focus on $\sqrt{(3+i)^3}$. We can simplify that into $(3+i)\sqrt{3+i}$. To simplify the square root, assume$$\sqrt{3+i}=a+bi\implies 3+i=a^2-b^2+2abi\tag3$$ Solving the system$$a^2-b^2=3$$$$ab=\dfrac 12$$ Gives two different values of $a,b$. Next, multiply them against $8i$ to get the answer.