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Natali5045456 [20]
3 years ago
6

Simplify the complex numbers using de Moivre's Theorem and match them with their solutions.

Mathematics
1 answer:
alexandr1967 [171]3 years ago
4 0

For the numbers in a+bi form, convert to polar form:


1+i=\sqrt2\dfrac{1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)

By DeMoivre's theorem,

(1+i)^5=(\sqrt2)^5\left(\cos\dfrac{5\pi}4+i\sin\dfrac{5\pi}4\right)=4\sqrt2\dfrac{-1-i}{\sqrt2}=-4-4i


-1+i=\sqrt2\dfrac{-1+i}{\sqrt2}=\sqrt2\left(\cos\dfrac{3\pi}4+i\sin\dfrac{3\pi}4}\right)

\implies(-1+i)^6=(\sqrt2)^6\left(\cos\dfrac{18\pi}4+i\sin\dfrac{18\pi}4\right)=8i


\sqrt3+i=2\dfrac{\sqrt3+i}2=2\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)

\implies2(\sqrt3+i)^{10}=2^{11}\left(\cos\dfrac{10\pi}6+i\sin\dfrac{10\pi}6\right)=2^{11}\dfrac{1-i\sqrt3}2=2^{10}(1-i\sqrt3)


For the numbers already in polar form, DeMoivre's theorem can be applied directly:


2\left(\cos20^\circ+i\sin20^\circ\right)^3=2\left(\cos60^\circ+i\sin60^\circ\right)=2\dfrac{1+i\sqrt3}2=1+i\sqrt3


2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)^4=2(\cos\pi+i\sin\pi)=-2


At second glance, I think the 2s in the last two numbers should also be getting raised to the 3rd and 4th powers:


\left(2(\cos20^\circ+i\sin20^\circ)\right)^3=8\left(\cos60^\circ+i\sin60^\circ\right)=4+4\sqrt3

\left(2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)\right)^4=16(\cos\pi+i\sin\pi)=-16

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