Question

Calcule (1+i)^10 e (1-i)^11.(sugestão calcule (1+i)^2 e (1-i)^2).

Answer 1

a)
$(1+i)^{10}=\\\\\left[(1+i)^2\right]^5=\\\\\left[1+2i+i^2\right]^5=\\\\(\cancel{1}+2i-\cancel{1})^5=\\\\(2i)^5=\\\\32i^5=\\\\32\cdot i^1=\\\\\boxed{32i}$

b)
$(1-i)^{11}=\\(1-i)^{10}\cdot(1-i)=\\\left[(1-i)^2\right]^5\cdot(1-i)=\\\left[1+2i+i^2\right]^5\cdot(1-i)=\\(\cancel{1}+2i-\cancel{1})^5\cdot(1-i)=\\(2i)^5\cdot(1-i)=\\32i^5\cdot(1-i)=\\32\cdot i^1\cdot(1-i)=\\32i(1-i)=\\32i-32i^2=\\\boxed{32i+32}$

Answer 2

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$\sf (1+i)^{10}\\\sf \rho=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\\\begin{cases}\sf cos(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\sf sen(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\end{cases}\implies\rm \theta=\dfrac{\pi}{4}\\\sf(1+i)^{10}=(\sqrt{2})^{10}\cdot\bigg[cos\bigg(10\cdot\dfrac{\pi}{4}\bigg)+i~sen\bigg(10\cdot\dfrac{\pi}{4}\bigg)\bigg]\\\sf (1+i)^{10}=32\cdot\bigg[cos\bigg(\dfrac{5\pi}{2}\bigg)+i~sen\bigg(\dfrac{5\pi}{2}\bigg)\bigg]\\\sf(1+i)^{10}=32\cdot(0+1)$

$\huge\boxed{\boxed{\boxed{\boxed{\sf(1+i)^{10}=32}}}}\blue{\checkmark}$