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# \left(2-2i\right)^6\cdot \frac{\left(\sqrt{3}+1\right)^3}{\left(\sqrt{3}+i\right)^4}

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(2-2i)^6(sqrt{3}+1)^3/(sqrt{3}+i)^4

Guest Aug 6, 2015

#1
+92751
+10

$$\\\dfrac{(2-2i)^6(\sqrt{3}+1)^3}{(\sqrt{3}+i)^4}\\\\\\ \dfrac{(2-2i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(\sqrt{3}+i)^4(\sqrt{3}-i)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{\{(\sqrt{3}+i)(\sqrt{3}-i)\}^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(3--1)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(4)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(2)^8}\\\\\\$$

$$\\\dfrac{(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{4}\\\\\\$$

The rest is just painful expansion.  Maybe you do not need to go any further.

(1-i)^6 will work out the easiest.  Use binomial expansions.

If you need help expanding I suggest you ask for it.   :)

Melody  Aug 7, 2015
#1
+92751
+10

$$\\\dfrac{(2-2i)^6(\sqrt{3}+1)^3}{(\sqrt{3}+i)^4}\\\\\\ \dfrac{(2-2i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(\sqrt{3}+i)^4(\sqrt{3}-i)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{\{(\sqrt{3}+i)(\sqrt{3}-i)\}^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(3--1)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(4)^4}\\\\\\ \dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(2)^8}\\\\\\$$

$$\\\dfrac{(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{4}\\\\\\$$

The rest is just painful expansion.  Maybe you do not need to go any further.

(1-i)^6 will work out the easiest.  Use binomial expansions.

If you need help expanding I suggest you ask for it.   :)

Melody  Aug 7, 2015