Show that: (1+isqrt3)(1+i)(cosvarphi+isinvarphi) =…
Show that: $$ (1+i\sqrt3)(1+i)(\cos\varphi+i\sin\varphi) = 2\sqrt2\left[\cos\left(\frac{7\pi}{12}+\varphi\right)+i\sin\left(\frac{7\pi}{12}+\varphi\right)\right]. $$
Explanation of Method
When complex numbers are multiplied, their moduli multiply and their arguments add: $r_1e^{i\theta_1}\cdot r_2e^{i\theta_2} = r_1r_2\, e^{i(\theta_1+\theta_2)}$. Converting each of the three factors to polar form reduces the whole product to multiplying two numbers and adding three angles.
Step 1 · Convert $1+i\sqrt3$ to polar form
$$ |1+i\sqrt3| = \sqrt{1^2+(\sqrt3)^2} = \sqrt4 = 2, \qquad \tan\theta_1 = \sqrt3 \implies \theta_1 = \frac{\pi}{3} $$
$$ 1+i\sqrt3 = 2\left(\cos\frac\pi3+i\sin\frac\pi3\right) = 2e^{i\pi/3} $$
Step 2 · Convert $1+i$ to polar form
$$ |1+i| = \sqrt{1^2+1^2} = \sqrt2, \qquad \tan\theta_2 = 1 \implies \theta_2 = \frac{\pi}{4} $$
$$ 1+i = \sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right) = \sqrt2\, e^{i\pi/4} $$
Step 3 · The third factor
$$ \cos\varphi+i\sin\varphi = e^{i\varphi}, \qquad \text{modulus } 1, \text{ argument } \varphi $$
Step 4 · Multiply in polar form
Multiplying all three: moduli multiply, arguments add.
Modulus: $$ 2\cdot\sqrt2\cdot1 = 2\sqrt2 $$
Argument: $$ \frac\pi3+\frac\pi4+\varphi = \frac{4\pi}{12}+\frac{3\pi}{12}+\varphi = \frac{7\pi}{12}+\varphi $$
Step 5 · Write the result
$$ (1+i\sqrt3)(1+i)(\cos\varphi+i\sin\varphi) = 2\sqrt2\, e^{i\left(\frac{7\pi}{12}+\varphi\right)} = \boxed{2\sqrt2\left[\cos\left(\frac{7\pi}{12}+\varphi\right)+i\sin\left(\frac{7\pi}{12}+\varphi\right)\right]} $$
which is exactly the identity to be shown. $\blacksquare$
Step 6 · Direct-multiplication check
As a cross-check, multiply the first two factors directly:
$$ (1+i\sqrt3)(1+i) = 1+i+i\sqrt3+i^2\sqrt3 = (1-\sqrt3)+i(1+\sqrt3) $$
Its modulus:
$$ \sqrt{(1-\sqrt3)^2+(1+\sqrt3)^2} = \sqrt{(1-2\sqrt3+3)+(1+2\sqrt3+3)} = \sqrt8 = 2\sqrt2 \quad \checkmark $$
matching Step 4's modulus. For the argument, $\tan\theta = \dfrac{1+\sqrt3}{1-\sqrt3}$; rationalizing gives $\tan\theta = -(2+\sqrt3)$. Independently, using $\frac{7\pi}{12}=\frac\pi3+\frac\pi4$ and the tangent addition formula on $\frac{5\pi}{12}=\frac\pi4+\frac\pi6$:
$$ \tan\frac{5\pi}{12} = \frac{\tan\frac\pi4+\tan\frac\pi6}{1-\tan\frac\pi4\tan\frac\pi6} = \frac{1+\frac1{\sqrt3}}{1-\frac1{\sqrt3}} = \frac{\sqrt3+1}{\sqrt3-1} = 2+\sqrt3 $$
and since $\frac{7\pi}{12}=\pi-\frac{5\pi}{12}$, $\tan\frac{7\pi}{12} = -\tan\frac{5\pi}{12} = -(2+\sqrt3)$, matching the direct calculation. ✓
Result
Proved by converting each factor to polar form ($2e^{i\pi/3}$, $\sqrt2 e^{i\pi/4}$, $e^{i\varphi}$), multiplying moduli ($2\sqrt2$) and adding arguments ($\frac{7\pi}{12}+\varphi$), then verified independently by direct multiplication.