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Exercise 4.1 · Q4

Q.Express the following in the form a+iba + ib: 3(7+i7)+i(7+i7)3(7 + i7) + i(7 + i7)

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The key idea is to treat ii as a number that obeys i2=−1i^2 = -1 and simplify the expression by distributing and combining like terms. The final result is 14+28i14 + 28i.

Concept and Intuition

Complex numbers are just numbers of the form a+iba + ib, where aa and bb are real numbers, and ii is the imaginary unit with the defining property i2=−1i^2 = -1. When you see an expression like 3(7+i7)+i(7+i7)3(7 + i7) + i(7 + i7), you're really just doing algebra you already know — distribution and combining like terms — with the extra rule that whenever you get an i2i^2, you replace it with −1-1.

The trick is to keep your real parts and imaginary parts separate. Think of ii as a label that tells you which "bucket" a term belongs to: terms without ii go into the real bucket, terms with ii go into the imaginary bucket. Once you've simplified everything, you write it as a+iba + ib.

Step-by-Step Solution

  1. Distribute the constants. Start with 3(7+i7)3(7 + i7). Multiply 33 by each term inside the parentheses:

3×7=21,3×i7=21i3 \times 7 = 21, \quad 3 \times i7 = 21i

So 3(7+i7)=21+21i3(7 + i7) = 21 + 21i.

  1. Distribute the ii in the second term. Now take i(7+i7)i(7 + i7). Multiply ii by each term:

i×7=7i,i×i7=7i2i \times 7 = 7i, \quad i \times i7 = 7i^2

So i(7+i7)=7i+7i2i(7 + i7) = 7i + 7i^2.

  1. Simplify using i2=−1i^2 = -1. Replace i2i^2 with −1-1:

7i2=7(−1)=−77i^2 = 7(-1) = -7

Therefore i(7+i7)=7i−7i(7 + i7) = 7i - 7.

  1. Combine the two results. Add the expressions from steps 1 and 3:

(21+21i)+(7i−7)(21 + 21i) + (7i - 7)

Group the real parts: 21+(−7)=1421 + (-7) = 14.

Group the imaginary parts: 21i+7i=28i21i + 7i = 28i. …

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