Title: Course Presentation

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Powers of Complex Numbers

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Some powers of the number \(i\):

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\(i^0=1,\qquad i^1=i,\qquad i^2=-1,\qquad i^3=-i,\dots\)

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The powers of the number \( i\) have four alternating values:

\(1,\qquad i,\qquad -1,\qquad -i,\qquad 1\)

\((a+bi)^2=(a+bi)\cdot(a+bi)=a^2+2abi+(bi)^2=a^2-b^2+2abi\)

\((a+bi)^3=(a+bi)^2\cdot(a+bi)=a^3+3a^2bi+3a(bi)^2+(bi)^3=\)

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\(=a^3-3ab^2+(3a^2-b^2)bi.\)

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Practice

Без названия Fill in the Blanks

Calculate.

Q: 1) \(i^5\) = [...].

2) \(i^0+i^4\) = [...].
A: i; 2
Feedback (0 to 100%): Lahendus: 1)\(i^5 = i^2\cdot i^2\cdot i = -1\cdot(-1)\cdot i = i\); 2) \(i^0+i^4 = 1+1 = 2\)

Multiplication Fill in the Blanks

Calculate.

Q: 1) \((2-i)^2=\)[...]

2) \((1-3i)^3=\) [...]
A: 3-4i; -26+18i
Feedback (0 to 100%): Solution: \(1) (2-i)^2 = 2^2-4i+i^2 = 4-1-4i = 3-4i;\) \(2)(1-3i)^3 \!=\! (1-3i)^2\cdot(1-3i) = (1^2-6i+(3i)^2)\cdot(1-3i) = (1-9-6i)\cdot(1-3i)=\) \(=(-8-6i)\cdot(1-3i) = -8\cdot1-8\cdot(-3i)-6i\cdot1-6i\cdot(-3i)=-8+24i-6i-18=\) \(= -26+18i\)