To calculate $\sqrt{-1 + i}$, we express the complex number in polar form and then apply De Moivre’s Theorem.

Step 1: Convert $-1 + i$ to Polar Form

A complex number $z = x + yi$ can be written in polar form as:

$z = r e^{i\theta}$

where:

• $r = |z| = \sqrt{x^2 + y^2}$
• $\theta = \tan^{-1} \left( \frac{y}{x} \right)$

For $z = -1 + i$:

$r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$

$\theta = \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1} (-1)$

Since $-1 + i$ is in the second quadrant, the angle is:

$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

Thus, we write:

$-1 + i = \sqrt{2} e^{i \frac{3\pi}{4}}$

Step 2: Compute the Square Root

The square roots of a complex number $r e^{i\theta}$ are given by:

$\sqrt{r} e^{i \frac{\theta}{2}}$

Substituting our values:

$\sqrt{\sqrt{2}} e^{i \frac{3\pi}{8}}$

Since $\sqrt{2} = 2^{1/2}$, we get:

$\sqrt{\sqrt{2}} = 2^{1/4}$

So the principal root is:

$2^{1/4} e^{i \frac{3\pi}{8}}$

Step 3: Convert Back to Rectangular Form

Using Euler’s formula:

$e^{i \frac{3\pi}{8}} = \cos \frac{3\pi}{8} + i \sin \frac{3\pi}{8}$

Approximating:

$\cos \frac{3\pi}{8} \approx 0.9239, \quad \sin \frac{3\pi}{8} \approx 0.3827$

Thus:

$2^{1/4} \cos \frac{3\pi}{8} + i 2^{1/4} \sin \frac{3\pi}{8}$

Approximating $2^{1/4} \approx 1.1892$:

$1.1892 (0.9239 + i 0.3827)$

$\approx 1.099 + 0.455 i$

Final Answer:

$\sqrt{-1 + i} = 2^{1/4} \cos \frac{3\pi}{8} + i 2^{1/4} \sin \frac{3\pi}{8}$

or approximately:

$1.099 + 0.455 i$

Would you like a more detailed explanation or verification?

Related Questions:

1. How do you find the cube root of a complex number?
2. What is De Moivre’s theorem and how is it used in roots of complex numbers?
3. Why do we convert complex numbers to polar form for root calculations?
4. What is the geometric interpretation of square roots of a complex number?
5. How does Euler’s formula relate to polar and rectangular forms of complex numbers?

Tip:

Always ensure your angle is in the correct quadrant when converting to polar form!