Question 16 of 28intermediate🔧 ApplyLong Answer4 marks

For any two complex numbers z1z_1 and z2z_2, prove that Re(z1z2)=Rez1Rez2−Imz1Imz2\text{Re}(z_1 z_2) = \text{Re} z_1 \text{Re} z_2 - \text{Im} z_1 \text{Im} z_2.

Correct Answer

Let z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, where aa, bb, cc, dd are real numbers.

Then Re(z1)=a\text{Re}(z_1) = a, Im(z1)=b\text{Im}(z_1) = b, Re(z2)=c\text{Re}(z_2) = c, and Im(z2)=d\text{Im}(z_2) = d.

Multiplying z1z_1 and z2z_2:

z1z2=(a+bi)(c+di)z_1 z_2 = (a + bi)(c + di)

Using the distributive law:

=a(c+di)+bi(c+di)= a(c + di) + bi(c + di) =ac+adi+bci+bdi2= ac + adi + bci + bdi^2

Since i2=−1i^2 = -1:

=ac+adi+bci−bd= ac + adi + bci - bd =(ac−bd)+(ad+bc)i= (ac - bd) + (ad + bc)i

Therefore, Re(z1z2)=ac−bd\text{Re}(z_1 z_2) = ac - bd.

Now,

Re z1⋅Re z2−Im z1⋅Im z2=a⋅c−b⋅d=ac−bd\text{Re}\, z_1 \cdot \text{Re}\, z_2 - \text{Im}\, z_1 \cdot \text{Im}\, z_2 = a \cdot c - b \cdot d = ac - bd

Hence, Re(z1z2)=Re z1⋅Re z2−Im z1⋅Im z2\text{Re}(z_1 z_2) = \text{Re}\, z_1 \cdot \text{Re}\, z_2 - \text{Im}\, z_1 \cdot \text{Im}\, z_2.

∴ Re(z1z2)=Re z1⋅Re z2−Im z1⋅Im z2\text{Re}(z_1 z_2) = \text{Re}\, z_1 \cdot \text{Re}\, z_2 - \text{Im}\, z_1 \cdot \text{Im}\, z_2

Exercise: Miscellaneous Exercise on Chapter 4 | Q: 2 | (Page 10)
For More Understanding

Explanation

This question tests the algebraic manipulation of complex numbers using their standard form a+bia + bi. The proof requires expressing both complex numbers in terms of their real and imaginary parts, then multiplying them using the distributive law (referenced in context under identities).

The key step is applying i2=−1i^2 = -1 to simplify the product. The result shows that the real part of a product equals the product of real parts minus the product of imaginary parts.

Solution Steps

  1. Step 1: Express z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di with real parts aa, cc and imaginary parts bb, dd

  2. Step 2: Multiply z1z_1 and z2z_2 using distributive law: (a+bi)(c+di)=ac+adi+bci+bdi2(a + bi)(c + di) = ac + adi + bci + bdi^2

  3. Step 3: Substitute i2=−1i^2 = -1 to get (ac−bd)+(ad+bc)i(ac - bd) + (ad + bc)i

  4. Step 4: Identify Re(z1z2)=ac−bd\text{Re}(z_1 z_2) = ac - bd

  5. Step 5: Calculate Re z1⋅Re z2−Im z1⋅Im z2=ac−bd\text{Re}\, z_1 \cdot \text{Re}\, z_2 - \text{Im}\, z_1 \cdot \text{Im}\, z_2 = ac - bd and conclude both sides are equal

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