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Exercise 4.1 · Q11

Q.Find the multiplicative inverse of 4−3i4 - 3i.

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The multiplicative inverse of a complex number zz is 1z\frac{1}{z}; we rationalize by multiplying numerator and denominator by the conjugate. The inverse of 4−3i4 - 3i is 425+325i\boxed{\frac{4}{25} + \frac{3}{25}i}.

The multiplicative inverse of a complex number zz is simply the number that, when multiplied by zz, gives 11. In other words, we need to find 14−3i\frac{1}{4-3i}.

The challenge is that we cannot leave a complex number in the denominator — we need to express our answer in standard form a+bia + bi. The key insight is to use the conjugate. When we multiply a complex number by its conjugate, the imaginary parts cancel and we get a real number. Specifically, (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2.

For 4−3i4 - 3i, the conjugate is 4+3i4 + 3i. Multiplying these gives (4−3i)(4+3i)=16+9=25(4-3i)(4+3i) = 16 + 9 = 25, a real number.

Here's how we find the inverse:

  1. Write the inverse as a fraction

    We want 14−3i\frac{1}{4-3i}.

  2. Multiply numerator and denominator by the conjugate

    The conjugate of 4−3i4 - 3i is 4+3i4 + 3i:

14−3i=14−3i⋅4+3i4+3i=4+3i(4−3i)(4+3i)\frac{1}{4-3i} = \frac{1}{4-3i} \cdot \frac{4+3i}{4+3i} = \frac{4+3i}{(4-3i)(4+3i)}

  1. Simplify the denominator

    Using the difference of squares pattern (or expanding directly): …

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