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mcr3u independent study unit – complex numbers assignment 1. simplify t…

Question

mcr3u independent study unit – complex numbers assignment

  1. simplify the following:

a) $i^{47}$
c) $i^{133}$
b) $i^{102}$
d) $i^{1212}$

  1. simplify the expression: $-21 + 26i + 45i^2 + 28i^5$. your answer must have a real part and an imaginary part.
  1. compute the following expressions.

a) $(46 + 26i) + (-25 + 67i)$
b) $(23 + 24i) - (26 + 46i)$
c) $(16 + 19i) + (-17 + 96i) - (75 - 46i)$

  1. find the product.

a) $(-3 - 24i)(27 + 8i)$
b) $(-14 - 32i)(7 - 23i)$

Explanation:

Question 1a: Simplify \( i^{47} \)

Step1: Recall \( i \) powers cycle

The powers of \( i \) cycle every 4: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), then repeat.
Divide 47 by 4: \( 47 \div 4 = 11 \) remainder \( 3 \).

Step2: Simplify using cycle

So \( i^{47} = i^{4 \times 11 + 3} = (i^4)^{11} \times i^3 \).
Since \( i^4 = 1 \), this becomes \( 1^{11} \times i^3 = i^3 \).
And \( i^3 = -i \).

Step1: Use \( i \) cycle

Divide 102 by 4: \( 102 \div 4 = 25 \) remainder \( 2 \).

Step2: Simplify

\( i^{102} = i^{4 \times 25 + 2} = (i^4)^{25} \times i^2 \).
\( i^4 = 1 \), so \( 1^{25} \times i^2 = i^2 \).
\( i^2 = -1 \).

Step1: Divide exponent by 4

\( 133 \div 4 = 33 \) remainder \( 1 \).

Step2: Simplify using cycle

\( i^{133} = i^{4 \times 33 + 1} = (i^4)^{33} \times i^1 \).
\( i^4 = 1 \), so \( 1^{33} \times i = i \).

Answer:

\( -i \)

Question 1b: Simplify \( i^{102} \)