Skip to content
Exercise 4.1 · Q12

Q.Find the multiplicative inverse of 5+3i\sqrt{5} + 3i.

Yanam BieapTextbookSubjective· 2mImportance★★★★★est
14% · 12/88 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The multiplicative inverse of a complex number zz is z‾∣z∣2\frac{\overline{z}}{|z|^2}. For 5+3i\sqrt{5} + 3i, this gives 5−3i14\frac{\sqrt{5} - 3i}{14}.

Why This Works

The multiplicative inverse of a complex number zz is another complex number z−1z^{-1} such that z⋅z−1=1z \cdot z^{-1} = 1. For real numbers, the inverse of aa is simply 1a\frac{1}{a}. But for complex numbers, we can't just write 15+3i\frac{1}{\sqrt{5} + 3i} — that's not in the standard form a+bia + bi.

The trick is to use the conjugate. When you multiply a complex number by its conjugate, you get a real number (the square of its modulus). This lets us "rationalise" the denominator, just like you do with surds.

For any complex number z=a+biz = a + bi,

z−1=z‾∣z∣2=a−bia2+b2z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}

Step-by-Step Solution

  1. Identify the number and its conjugate.

    We have z=5+3iz = \sqrt{5} + 3i. Its conjugate is z‾=5−3i\overline{z} = \sqrt{5} - 3i.

  2. Find the modulus squared.

    ∣z∣2=(5)2+(3)2=5+9=14|z|^2 = (\sqrt{5})^2 + (3)^2 = 5 + 9 = 14.

  3. Apply the formula.

z−1=z‾∣z∣2=5−3i14z^{-1} = \frac{\overline{z}}{|z|^2} = \frac{\sqrt{5} - 3i}{14}

  1. Write in standard form. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.