Question

Show that 2|ψ1||ψ2| cos (φ1 − φ2) = 2 (A1A2 + B1B2)
given that ψ1 = A1 + iB1 and ψ2 = A2 + iB2.

Answer 1

$\textbf{Concept used:}$

$\text{Let }\;z=x+i\,y$

$\text{Then, Modulus of z is}\bf\;|z|=\sqrt{x^2+y^2}$

$\text{and argument of z is}\bf\;tan\theta=\frac{y}{x}$

$\text{Here,}\;cos\theta=\frac{x}{|z|}\;\text{and}\;sin\theta=\frac{y}{|z|}$

$\textbf{Given:}$

${\psi}_1=A_1+i\,B_1\;\text{and}\;{\psi}_2=A_2+i\,B_2$

$\textbf{To prove:}$

$2\,|{\psi}_1|\,|{\psi}_2|\,cos({\varphi}_1-{\varphi}_2)=2(A_1\,A_2+B_1\,B_2)$

${\psi}_1=A_1+i\,B_1\;\text{and}\;{\psi}_2=A_2+i\,B_2$

$\text{Then,}$

$|{\psi}_1|=\sqrt{{A_1}^2+{B_1}^2}\;\text{and}\;|{\psi}_2|=\sqrt{{A_2}^2+{B_2}^2}$

$cos{\varphi}_1=\dfrac{A_1}{\sqrt{{A_1}^2+{B_1}^2}}\;\text{and}\;sin{\varphi}_1=\dfrac{B_1}{\sqrt{{A_1}^2+{B_1}^2}}$

$cos{\varphi}_2=\dfrac{A_2}{\sqrt{{A_2}^2+{B_2}^2}}\;\text{and}\;sin{\varphi}_2=\dfrac{B_2}{\sqrt{{A_2}^2+{B_2}^2}}$

$\text{Consider,}$

$cos({\varphi}_1-{\varphi}_2)=cos{\varphi}_1\,cos{\varphi}_2+sin{\varphi}_1\,sin{\varphi}_2$

$\implies\,cos({\varphi}_1-{\varphi}_2)=\dfrac{A_1}{\sqrt{{A_1}^2+{B_1}^2}}\,\dfrac{A_2}{\sqrt{{A_2}^2+{B_2}^2}}+\dfrac{B_1}{\sqrt{{A_1}^2+{B_1}^2}}\,\dfrac{B_2}{\sqrt{{A_2}^2+{B_2}^2}}$

$\implies\,cos({\varphi}_1-{\varphi}_2)=\dfrac{A_1\,A_2}{\sqrt{{A_1}^2+{B_1}^2}\,\sqrt{{A_2}^2+{B_2}^2}}+\dfrac{B_1\,B_2}{\sqrt{{A_1}^2+{B_1}^2}\,\sqrt{{A_2}^2+{B_2}^2}}$

$\implies\,cos({\varphi}_1-{\varphi}_2)=\dfrac{A_1\,A_2+B_1\,B_2}{\sqrt{{A_1}^2+{B_1}^2}\,\sqrt{{A_2}^2+{B_2}^2}}$

$\implies\,cos({\varphi}_1-{\varphi}_2)=\dfrac{A_1\,A_2+B_1\,B_2}{|{\psi}_1|\,|{\psi}_2|}$

$\implies|{\psi}_1|\,|{\psi}_2|\,cos({\varphi}_1-{\varphi}_2)=(A_1\,A_2+B_1\,B_2)$

$\therefore\bf\,2|{\psi}_1|\,|{\psi}_2|\,cos({\varphi}_1-{\varphi}_2)=2(A_1\,A_2+B_1\,B_2)$