Question Number 226581 by klipto last updated on 06/Dec/25
$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{polar}}\:\boldsymbol{\mathrm{of}} \\ $$$$\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\sqrt{\mathrm{3}}\right) \\ $$
Answered by Frix last updated on 06/Dec/25
$$=\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} ×\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} =\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{7}\pi}{\mathrm{12}}} \\ $$
Answered by Kademi last updated on 06/Dec/25
$$\left(\:\sqrt{\mathrm{2}}{e}^{\frac{\pi\mathrm{i}}{\mathrm{4}}} \right)\left(\mathrm{2}{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)\:=\:\mathrm{2}\sqrt{\mathrm{2}}{e}^{\frac{\mathrm{7}\pi\mathrm{i}}{\mathrm{12}}} \\ $$
Answered by Spillover last updated on 06/Dec/25
![Z=∣z∣[cos θ+isin θ] Z=∣z_1 ∣∣z_2 ∣ z_1 =1+i ∣z_1 ∣=(√2) tan θ=1 θ=(π/4) z_2 =1+i(√(3 )) ∣z_2 ∣=2 tan θ=(√3) θ=tan^(−1) (√3) =(π/3) Z=∣z_1 ∣∣z_2 ∣ Z=∣z∣[cos θ+isin θ] Z_1 =(√(2 ))[cos (π/4)+isin (π/4)] Z_2 =2[cos (π/3)+isin (π/3)] Z_1 Z_2 =2(√2)[cos ((7π)/(12))+isin ((7π)/(12))]](https://www.tinkutara.com/question/Q226590.png)
$${Z}=\mid{z}\mid\left[\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right] \\ $$$${Z}=\mid{z}_{\mathrm{1}} \mid\mid{z}_{\mathrm{2}} \mid \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\:\:\:\:\mid{z}_{\mathrm{1}} \mid=\sqrt{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$${z}_{\mathrm{2}} =\mathrm{1}+{i}\sqrt{\mathrm{3}\:}\:\:\:\:\:\:\mid{z}_{\mathrm{2}} \mid=\mathrm{2} \\ $$$$\mathrm{tan}\:\theta=\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{3}}\:=\frac{\pi}{\mathrm{3}} \\ $$$${Z}=\mid{z}_{\mathrm{1}} \mid\mid{z}_{\mathrm{2}} \mid \\ $$$${Z}=\mid{z}\mid\left[\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right] \\ $$$${Z}_{\mathrm{1}} =\sqrt{\mathrm{2}\:}\left[\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right] \\ $$$${Z}_{\mathrm{2}} =\mathrm{2}\left[\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right] \\ $$$${Z}_{\mathrm{1}} {Z}_{\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{12}}+{i}\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{12}}\right] \\ $$