Suppose z1 = r1(cos θ1 + isin θ1 ) and z2 = r2(cos θ2 + isin θ2). Prove: (a) z1z2 = r1r2{cos(θ1 + θ2) + isin(θ1 + θ2)}, (b) z1/z2= r1/r2{cos(θ1 - θ2) + isin(θ1 - θ2)}

Question

Suppose [latex] z_1=r_1\left(\cos heta_1+i \sin heta_1 ight) ext { and } z_2=r_2\left(\cos heta_2+i \sin heta_2 ight) [/latex]. Prove:

(a) [latex]z_1 z_2=r_1 r_2\left\{\cos \left( heta_1+ heta_2 ight)+i \sin \left( heta_1+ heta_2 ight) ight\}[/latex] (b) [latex]\frac{z_1}{z_2}=\frac{r_1}{r_2}\left\{\cos \left( heta_1- heta_2 ight)+i \sin \left( heta_1- heta_2 ight) ight\}[/latex].

Accepted Answer

[latex] \begin{aligned} (a) z_1 z_2 &=\left\{r_1\left(\cos heta_1+i \sin heta_1 ight) ight\}\left\{r_2\left(\cos heta_2+i \sin heta_2 ight) ight\} \ &=r_1 r_2\left\{\left(\cos heta_1 \cos heta_2-\sin heta_1 \sin heta_2 ight)+i\left(\sin heta_1 \cos heta_2+\cos heta_1 \sin heta_2 ight) ight\} \ &=r_1 r_2\left\{\cos \left( heta_1+ heta_2 ight)+i \sin \left( heta_1+ heta_2 ight) ight\} \end{aligned} [/latex]

[latex] \begin{aligned} (b) \frac{z_1}{z_2} &=\frac{r_1\left(\cos heta_1+i \sin heta_1 ight)}{r_2\left(\cos heta_2+i \sin heta_2 ight)} \cdot \frac{\left(\cos heta_2-i \sin heta_2 ight)}{\left(\cos heta_2-i \sin heta_2 ight)} \ &=\frac{r_1}{r_2}\left\{\frac{\left(\cos heta_1 \cos heta_2+\sin heta_1 \sin heta_2 ight)+i\left(\sin heta_1 \cos heta_2-\cos heta_1 \sin heta_2 ight)}{\cos ^2 heta_2+\sin ^2 heta_2} ight\} \ &=\frac{r_1}{r_2}\left\{\cos \left( heta_1- heta_2 ight)+i \sin \left( heta_1- heta_2 ight) ight\} \end{aligned} [/latex]

In terms of Euler's formula, [latex]e^{i heta}=\cos heta+i \sin heta[/latex], the results state that if [latex]z_1=r_1 e^{i heta_1} and z_2=r_2 e^{i heta_2}[/latex], then [latex]z_1 z_2=r_1 r_2 e^{i\left( heta_1+ heta_2 ight)}[/latex] and [latex]z_1 / z_2=r_1 e^{i heta_1} / r_2 e^{i heta_2}=\left(r_1 / r_2 ight) e^{i\left( heta_1- heta_2 ight)}[/latex].

Question 1.19: Suppose z1 = r1(cos θ1 + isin θ1 ) and z2 = r2(cos θ2 + isin...

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