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1. simplify the following: a) $i^{47}$ $=i^{4×11 + 3}$ $=(i^4)^{11}×i^3…

Question

  1. simplify the following:

a) $i^{47}$
$=i^{4×11 + 3}$
$=(i^4)^{11}×i^3$
$=-i$
b) $i^{102}$
=
c) $i^{133}$
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:

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

Step 1: Recall the cycle of \( i \)

The powers of \( i \) cycle every 4: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and then repeat. So we divide the exponent by 4 and find the remainder.
\( 47 \div 4 = 11 \) with a remainder of \( 3 \). So \( i^{47} = i^{4 \times 11 + 3} = (i^4)^{11} \times i^3 \)

Step 2: Simplify using the cycle

Since \( i^4 = 1 \) and \( i^3 = -i \), we have \( (1)^{11} \times (-i) = -i \)

Step 1: Find the remainder when 102 is divided by 4

\( 102 \div 4 = 25 \) with a remainder of \( 2 \). So \( i^{102} = i^{4 \times 25 + 2} = (i^4)^{25} \times i^2 \)

Step 2: Simplify using the cycle

Since \( i^4 = 1 \) and \( i^2 = -1 \), we have \( (1)^{25} \times (-1) = -1 \)

Step 1: Find the remainder when 133 is divided by 4

\( 133 \div 4 = 33 \) with a remainder of \( 1 \). So \( i^{133} = i^{4 \times 33 + 1} = (i^4)^{33} \times i^1 \)

Step 2: Simplify using the cycle

Since \( i^4 = 1 \) and \( i^1 = i \), we have \( (1)^{33} \times i = i \)

Answer:

\( -i \)

Part b: Simplify \( i^{102} \)