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The problem asks us to find the quotient of the given complex numbers and express the answer in the standard form $a + bi$, where $a$ and $b$ are real numbers. We have six parts: a. $\frac{4i}{3-2i}$ b. $\frac{2+5i}{3+7i}$ c. $\frac{4-10i}{-3+7i}$ d. $\frac{-1-i}{-2-3i}$ e. $\frac{3+9i}{4-i}$ f. $\frac{-4+9i}{-3-6i}$

AlgebraComplex NumbersComplex Number DivisionComplex Conjugate
2025/4/9

1. Problem Description

The problem asks us to find the quotient of the given complex numbers and express the answer in the standard form a+bia + bi, where aa and bb are real numbers. We have six parts:
a. 4i3−2i\frac{4i}{3-2i}
b. 2+5i3+7i\frac{2+5i}{3+7i}
c. 4−10i−3+7i\frac{4-10i}{-3+7i}
d. −1−i−2−3i\frac{-1-i}{-2-3i}
e. 3+9i4−i\frac{3+9i}{4-i}
f. −4+9i−3−6i\frac{-4+9i}{-3-6i}

2. Solution Steps

To divide complex numbers, we multiply the numerator and denominator by the conjugate of the denominator.
a. 4i3−2i=4i(3+2i)(3−2i)(3+2i)=12i+8i29−(4i2)=12i−89+4=−8+12i13=−813+1213i\frac{4i}{3-2i} = \frac{4i(3+2i)}{(3-2i)(3+2i)} = \frac{12i+8i^2}{9 - (4i^2)} = \frac{12i - 8}{9+4} = \frac{-8+12i}{13} = -\frac{8}{13} + \frac{12}{13}i
b. 2+5i3+7i=(2+5i)(3−7i)(3+7i)(3−7i)=6−14i+15i−35i29−49i2=6+i+359+49=41+i58=4158+158i\frac{2+5i}{3+7i} = \frac{(2+5i)(3-7i)}{(3+7i)(3-7i)} = \frac{6-14i+15i-35i^2}{9 - 49i^2} = \frac{6+i+35}{9+49} = \frac{41+i}{58} = \frac{41}{58} + \frac{1}{58}i
c. 4−10i−3+7i=(4−10i)(−3−7i)(−3+7i)(−3−7i)=−12−28i+30i+70i29−49i2=−12+2i−709+49=−82+2i58=−41+i29=−4129+129i\frac{4-10i}{-3+7i} = \frac{(4-10i)(-3-7i)}{(-3+7i)(-3-7i)} = \frac{-12-28i+30i+70i^2}{9 - 49i^2} = \frac{-12+2i-70}{9+49} = \frac{-82+2i}{58} = \frac{-41+i}{29} = -\frac{41}{29} + \frac{1}{29}i
d. −1−i−2−3i=(−1−i)(−2+3i)(−2−3i)(−2+3i)=2−3i+2i−3i24−9i2=2−i+34+9=5−i13=513−113i\frac{-1-i}{-2-3i} = \frac{(-1-i)(-2+3i)}{(-2-3i)(-2+3i)} = \frac{2-3i+2i-3i^2}{4-9i^2} = \frac{2-i+3}{4+9} = \frac{5-i}{13} = \frac{5}{13} - \frac{1}{13}i
e. 3+9i4−i=(3+9i)(4+i)(4−i)(4+i)=12+3i+36i+9i216−i2=12+39i−916+1=3+39i17=317+3917i\frac{3+9i}{4-i} = \frac{(3+9i)(4+i)}{(4-i)(4+i)} = \frac{12+3i+36i+9i^2}{16 - i^2} = \frac{12+39i-9}{16+1} = \frac{3+39i}{17} = \frac{3}{17} + \frac{39}{17}i
f. −4+9i−3−6i=(−4+9i)(−3+6i)(−3−6i)(−3+6i)=12−24i−27i+54i29−36i2=12−51i−549+36=−42−51i45=−14−17i15=−1415−1715i\frac{-4+9i}{-3-6i} = \frac{(-4+9i)(-3+6i)}{(-3-6i)(-3+6i)} = \frac{12-24i-27i+54i^2}{9 - 36i^2} = \frac{12-51i-54}{9+36} = \frac{-42-51i}{45} = \frac{-14-17i}{15} = -\frac{14}{15} - \frac{17}{15}i

3. Final Answer

a. −813+1213i-\frac{8}{13} + \frac{12}{13}i
b. 4158+158i\frac{41}{58} + \frac{1}{58}i
c. −4129+129i-\frac{41}{29} + \frac{1}{29}i
d. 513−113i\frac{5}{13} - \frac{1}{13}i
e. 317+3917i\frac{3}{17} + \frac{39}{17}i
f. −1415−1715i-\frac{14}{15} - \frac{17}{15}i

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