Question

If z1=3+5i,z2=2-3i, then prove that
Conjugate of (z1/z2)= conjugate of z1/ conjugate of z2

Answer 1

Answer:

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Answer 2

Answer:

Given:

• $z_1=3+5i$.

• $z_2=2-3i$.

where,

$i=\sqrt{-1}$.

To prove:

$\text{Conjugate of } \left(\dfrac {z_1}{z_2}\right )=\dfrac{\text{Conjugate of }z_1}{\text{Conjugate of }z_2}.$

Proof:

LHS:

$\dfrac{z_1}{z_2} =\dfrac{3+5i}{2-3i}\\=\dfrac{3+5i}{2-3i}\times \dfrac{2+3i}{2+3i}\ \ \ \ \text{(On rationalizing)}\\=\dfrac{(3+5i)\times (2+3i)}{(2-3i)\times (2+3i)}\\=\dfrac{6+9i+10i+15i^2}{4+6i-6i-9i^2}\\=\dfrac{6-15+19i}{4+9}\ \ \ \ \ \ \ \because i^2=-1\\=\dfrac{-9+19i}{13}.$

Therefore,

$\text{Conjugate of } \left(\dfrac {z_1}{z_2}\right )=\text{Conjugate of }\left(\dfrac{-9+19i}{13}\right)=\left(\dfrac{-9-19i}{13}\right).$

RHS:

$\text{Conjugate of } z_1 = 3-5i\\\text{Conjugate of } z_2 = 2+3i\\\dfrac{\text{Conjugate of } z_1}{\text{Conjugate of } z_2}=\dfrac{3-5i}{2+3i}\\=\dfrac{3-5i}{2+3i}\times \dfrac{2-3i}{2-3i}\ \ \ \ \text{(On rationalizing)}\\=\dfrac{(3-5i)\times (2-3i)}{(2+3i)\times (2-3i)}\\=\dfrac{6-9i-10i+15i^2}{4-6i+6i-9i^2}\\=\dfrac{6-15-19i}{4+9}\ \ \ \ \ \ \because i^2=-1.\\=\dfrac{-9-19i}{13}.$

Thus, LHS = RHS, proved.