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Miscellaneous Exercise · Q3

Q.Reduce (11−4i−21+i)(3−4i5+i)\left(\dfrac{1}{1 - 4i} - \dfrac{2}{1 + i}\right)\left(\dfrac{3 - 4i}{5 + i}\right) to the standard form.

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Combine the first bracket over a common denominator, multiply by the second fraction, then rationalise. The standard form is 307442+599442i\dfrac{307}{442} + \dfrac{599}{442}i.

Combine the first bracket. Over the common denominator (1−4i)(1+i)(1-4i)(1+i):

11−4i−21+i=(1+i)−2(1−4i)(1−4i)(1+i)=−1+9i5−3i,\frac{1}{1-4i}-\frac{2}{1+i} = \frac{(1+i)-2(1-4i)}{(1-4i)(1+i)} = \frac{-1+9i}{5-3i},

since (1−4i)(1+i)=1−3i−4i2=5−3i.(1-4i)(1+i)=1-3i-4i^2 = 5-3i.

Multiply by the second fraction.

−1+9i5−3i⋅3−4i5+i=(−1+9i)(3−4i)(5−3i)(5+i)=33+31i28−10i,\frac{-1+9i}{5-3i}\cdot\frac{3-4i}{5+i} = \frac{(-1+9i)(3-4i)}{(5-3i)(5+i)} = \frac{33+31i}{28-10i},

because (−1+9i)(3−4i)=−3+31i−36i2=33+31i(-1+9i)(3-4i) = -3+31i-36i^2 = 33+31i and (5−3i)(5+i)=25−10i−3i2=28−10i.(5-3i)(5+i) = 25-10i-3i^2 = 28-10i.

Rationalise by multiplying by the conjugate 28+10i28+10i:

(33+31i)(28+10i)282+102=614+1198i884,\frac{(33+31i)(28+10i)}{28^2+10^2} = \frac{614+1198i}{884},

since (33+31i)(28+10i)=924+1198i+310i2=614+1198i(33+31i)(28+10i) = 924+1198i+310i^2 = 614+1198i and 282+102=884.28^2+10^2 = 884.

Reduce by dividing numerator and denominator by 22:

307+599i442=307442+599442i.\frac{307+599i}{442} = \frac{307}{442} + \frac{599}{442}i.

(307307 and 599599 are prime and 442=2⋅13⋅17442=2\cdot 13\cdot 17, so this is in lowest terms.)

✓Final answer

The expression in standard form is 307442+599442i.\dfrac{307}{442} + \dfrac{599}{442}i.

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