Simplify complex exponential

Question

This is a solution from a textbook.
I confirmed with an online calculator that this solution is true.

$$ 1 - e^{-j*0.4\pi} = 1.176e^{j*0.3\pi}$$

How do you get from the left hand side to the right hand side?

Answer 1

First, convert what you are given on the LHS to polar coordinates. We have that

$$e^{-i\theta}=\cos{\theta}-i\sin{\theta}$$

therefore

\begin{align}e^{-i(0.4)\pi}=\cos(0.4\pi)-i\sin(0.4\pi)&=\cos(72^{\circ})-i\sin(72^{\circ})\\&=\frac{1}{4} (\sqrt{5} - 1)-i\Big(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}\Big)\\&\approx0.309017-0.951057i\end{align}

and

$$1-e^{-i(0.4)\pi}\approx 1-\big(0.309017-0.951057i\big) = 0.690983+0.951057i$$

which means that

$$x\approx 0.690983, y \approx 0.951057$$

so in the conversion to polar coordinates we can find the radius $r$ through

$$x^2 + y^2 =r^2 \implies r=\sqrt{x^2+y^2} \implies r=\sqrt{0.690983^2 + 0.951057^2} \approx 1.17557$$

and the angle $\theta$ by

$$\tan\theta = \frac{y}{x} \implies \theta = \arctan\Big(\frac{y}{x}\Big)\implies \theta \approx\arctan\Big(\frac{0.951057}{0.690983}\Big)= 54^{\circ}$$

which gives us the polar coordinates as

$$r \approx 1.17557,~~\theta=54^{\circ} $$

we can then convert from polar form to complex form by observing that

$$e^{i\theta}=\cos\theta + i\sin\theta$$

produces the RHS since

$$1.17557(\cos{54^{\circ}}+i\sin{54^{\circ}})=1.17557(\cos{0.3\pi}+i\sin{0.3\pi})=1.17557e^{i(0.3\pi)}$$

and so the RHS is therefore

$$re^{i\theta}$$

Answer 2

Express the exponential on LHS in rectangular form. Perform the addition and change the sum again to polar form.