Skip to content
Exercise 1.2 · Q1

Q.Find the complex conjugates and modulus of the following complex numbers: 1−i1 - i, 10+4i10 + 4i, (3+5i)(4+6i)(3 + 5i)(4 + 6i), 2+7i5+4i\dfrac{2 + 7i}{5 + 4i}.

Dnh Dd CbseNCERTSubjective· 3mImportance★★★★★est
28% · 5/18 Questions
✓ Free question

Each complex number is conjugated by flipping the sign of its imaginary part, and its modulus is a2+b2\sqrt{a^2+b^2}; for the product/quotient the number is simplified first, then the same rules applied.

[!FORMULA]

For z=a+biz=a+bi: conjugate zˉ=a−bi\bar z=a-bi; modulus ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}. Also ∣zw∣=∣z∣ ∣w∣|zw|=|z|\,|w| and ∣zw∣=∣z∣∣w∣\left|\dfrac{z}{w}\right|=\dfrac{|z|}{|w|}.

  1. z=1−iz=1-i: zˉ=1+i\bar z=1+i; ∣z∣=12+(−1)2=2≈1.414|z|=\sqrt{1^2+(-1)^2}=\sqrt2\approx1.414.
  2. z=10+4iz=10+4i: zˉ=10−4i\bar z=10-4i; ∣z∣=100+16=116=229≈10.770|z|=\sqrt{100+16}=\sqrt{116}=2\sqrt{29}\approx10.770.
  3. z=(3+5i)(4+6i)z=(3+5i)(4+6i): expand =12+18i+20i+30i2=12+38i−30=−18+38i=12+18i+20i+30i^2=12+38i-30=-18+38i. So zˉ=−18−38i\bar z=-18-38i; ∣z∣=182+382=324+1444=1768=2442≈42.048|z|=\sqrt{18^2+38^2}=\sqrt{324+1444}=\sqrt{1768}=2\sqrt{442}\approx42.048. Check: ∣z∣=∣3+5i∣∣4+6i∣=3452=1768|z|=|3+5i||4+6i|=\sqrt{34}\sqrt{52}=\sqrt{1768} — matches.
  4. z=2+7i5+4iz=\dfrac{2+7i}{5+4i}: multiply by 5−4i5−4i\dfrac{5-4i}{5-4i}: numerator (2+7i)(5−4i)=10−8i+35i−28i2=10+28+27i=38+27i(2+7i)(5-4i)=10-8i+35i-28i^2=10+28+27i=38+27i; denominator 25+16=4125+16=41. So z=3841+2741iz=\dfrac{38}{41}+\dfrac{27}{41}i. zˉ=3841−2741i\bar z=\dfrac{38}{41}-\dfrac{27}{41}i; ∣z∣=382+27241=1444+72941=217341≈1.137|z|=\dfrac{\sqrt{38^2+27^2}}{41}=\dfrac{\sqrt{1444+729}}{41}=\dfrac{\sqrt{2173}}{41}\approx1.137. Check: ∣z∣=∣2+7i∣∣5+4i∣=5341=53/41≈1.137|z|=\dfrac{|2+7i|}{|5+4i|}=\dfrac{\sqrt{53}}{\sqrt{41}}=\sqrt{53/41}\approx1.137 — matches.
✓Final answer

1−i→zˉ=1+i, ∣z∣=21-i\to \bar z=1+i,\ |z|=\sqrt2.  10+4i→zˉ=10−4i, ∣z∣=229\ 10+4i\to \bar z=10-4i,\ |z|=2\sqrt{29}.  (3+5i)(4+6i)=−18+38i→zˉ=−18−38i, ∣z∣=2442\ (3+5i)(4+6i)=-18+38i\to \bar z=-18-38i,\ |z|=2\sqrt{442}.  2+7i5+4i=3841+2741i→zˉ=3841−2741i, ∣z∣=217341\ \dfrac{2+7i}{5+4i}=\dfrac{38}{41}+\dfrac{27}{41}i\to \bar z=\dfrac{38}{41}-\dfrac{27}{41}i,\ |z|=\dfrac{\sqrt{2173}}{41}.

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.