Question

The polar form of the complex number \(z = \frac{1}{1+i}\), (where \(i = \sqrt{-1}\)) is

• A. \(\frac{1}{\sqrt{2}}(cos\frac{7π}{4}+isin\frac{7π}{4})\)
• B. \(\frac{1}{\sqrt{2}}(cos\frac{π}{4}+isin\frac{π}{4})\)
• C. \(\frac{1}{2}(cos\frac{π}{4}+isin\frac{π}{4})\)
• D. \(\frac{1}{2}(cos\frac{7π}{4}+isin\frac{7π}{4})\)

Hint:

Rationalise to get the real and imaginary parts, then find modulus and argument, remembering the quadrant.

Answer 1

The Correct Option is A

Step 1: Use properties of modulus and argument.
$|1/(1+i)| = 1/|1+i| = 1/\sqrt{2}$.

Step 2: Argument.
$\arg(1/w) = -\arg(w)$. Since $\arg(1+i) = \pi/4$, we get $\arg z = -\pi/4$, which is the same as $7\pi/4$ in $[0, 2\pi)$.

Step 3: Assemble.
$z = \dfrac{1}{\sqrt{2}}\left(\cos\dfrac{7\pi}{4} + i\sin\dfrac{7\pi}{4}\right)$. Check: $\cos(7\pi/4) = 1/\sqrt{2}$, $\sin(7\pi/4) = -1/\sqrt{2}$, giving $\tfrac{1}{2} - \tfrac{1}{2}i$, as expected.

Final Answer:
Option (A) is correct. \[ \boxed{\frac{1}{\sqrt{2}}\left(\cos\frac{7\pi}{4} + i\sin\frac{7\pi}{4}\right)} \]