Question

which expressions are equivalent to the given complex number? 45 + 2i (9 + 4i) + 2(4 + 7i)(1 − 2i) (2 + 8i) + (30 − 6i) (13 + 4i) + (32 − 6i) (2 + 8i) + 2(9 + 6i)(1−i) (9 + 4i) + (36 − 2i) (13 + 4i) + 2(7 + 2i)(2−i)

Explanation:

Step1: Evaluate first expression

First expand $(4+7i)(1-2i)$:
$$(4+7i)(1-2i)=4(1)+4(-2i)+7i(1)+7i(-2i)=4-8i+7i-14i^2$$
Since $i^2=-1$, substitute:
$$4 - i -14(-1)=4 - i +14=18 - i$$
Multiply by 2: $2(18 - i)=36 - 2i$
Add $(9+4i)$:
$$(9+4i)+(36-2i)=(9+36)+(4i-2i)=45+2i$$

Step2: Evaluate second expression

Add real and imaginary parts separately:
$$(2+30)+(8i-6i)=32+2i$$

Step3: Evaluate third expression

Add real and imaginary parts separately:
$$(13+32)+(4i-6i)=45-2i$$

Step4: Evaluate fourth expression

First expand $(9+6i)(1-i)$:
$$(9+6i)(1-i)=9(1)+9(-i)+6i(1)+6i(-i)=9-9i+6i-6i^2$$
Since $i^2=-1$, substitute:
$$9 - 3i -6(-1)=9 - 3i +6=15 - 3i$$
Multiply by 2: $2(15 - 3i)=30 - 6i$
Add $(2+8i)$:
$$(2+8i)+(30-6i)=(2+30)+(8i-6i)=32+2i$$

Step5: Evaluate fifth expression

Add real and imaginary parts separately:
$$(9+36)+(4i-2i)=45+2i$$

Step6: Evaluate sixth expression

First expand $(7+2i)(2-i)$:
$$(7+2i)(2-i)=7(2)+7(-i)+2i(2)+2i(-i)=14-7i+4i-2i^2$$
Since $i^2=-1$, substitute:
$$14 - 3i -2(-1)=14 - 3i +2=16 - 3i$$
Multiply by 2: $2(16 - 3i)=32 - 6i$
Add $(13+4i)$:
$$(13+4i)+(32-6i)=(13+32)+(4i-6i)=45-2i$$

Answer:

• $(9 + 4i) + 2(4 + 7i)(1 - 2i)$
• $(9 + 4i) + (36 - 2i)$