Expand $(\cos x+i\sin x)^2$ and solve $(\cos x+i\sin x)^2=1$

Expert intro: This problem links complex numbers with trigonometric identities. Expanding the expression and comparing real and imaginary parts gives the required solutions cleanly.

Detailed walkthrough

Step 1: Expand the square

Let

$z=\cos x+i\sin x.$

Then

$z^2=(\cos x+i\sin x)^2.$

Expanding gives

$\cos^2 x-\sin^2 x+2i\sin x\cos x.$

So the equation

$(\cos x+i\sin x)^2=1$

becomes

$\cos^2 x-\sin^2 x+2i\sin x\cos x=1.$

Step 2: Match real and imaginary parts

Since $1$ is real, the imaginary part must be zero:

$2\sin x\cos x=0.$

So

$\sin x=0 \quad \text{or} \quad \cos x=0.$

Case 1: $\sin x=0$

For $0\le x\le \pi$, this gives

$x=0,\ \pi.$

Check the original equation:

• $x=0$: $(1+i\cdot 0)^2=1$
• $x=\pi$: $(-1+i\cdot 0)^2=1$

Both work.

Case 2: $\cos x=0$

For $0\le x\le \pi$, this gives

$x=\frac{\pi}{2}.$

Check:

$(0+i\cdot 1)^2=-1,$

so this does not satisfy the equation.

Step 3: List the solutions

Thus the solutions of

$(\cos x+i\sin x)^2=1$

for $0\le x\le \pi$ are

$x=0,\ \pi.$

Step 4: Verify the trigonometric equation

Now consider

$\cos x+\sin x=1.$

Rewrite as

$\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)=1,$

so

$\cos\left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt2}.$

For $0\le x\le \pi$, the solutions are

$x=0 \quad \text{and} \quad x=\frac{\pi}{2}.$

Final answer

• $(\cos x+i\sin x)^2=1$ on $[0,\pi]$ gives $x=0,\pi$.
• $\cos x+\sin x=1$ on $[0,\pi]$ has solutions $x=0$ and $x=\frac{\pi}{2}$.

💡 Pitfall guide

Do not stop after setting the imaginary part to zero; you must also check the real part and verify each candidate in the original equation. Also, $x=\frac{\pi}{2}$ solves $\cos x+\sin x=1$ but not $(\cos x+i\sin x)^2=1$.

🔄 Real-world variant

If the equation were $(\cos x+i\sin x)^n=1$, then De Moivre’s theorem would give $nx=2k\pi$ for integers $k$. The solution set on a restricted interval would depend on $n$ and the chosen domain.

🔍 Related terms

De Moivre’s theorem, complex roots, trigonometric identity