We need to compute:

$(-3 + 2i)^5$

Step 1: Convert to Polar Form

A complex number $a + bi$ can be expressed in polar form as:

$r e^{i\theta}$

where:

• $r = |a + bi| = \sqrt{a^2 + b^2}$
• $\theta = \arg(a + bi) = \tan^{-1} \left(\frac{b}{a} \right)$

For $-3 + 2i$:

$r = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}$

The argument $\theta$ is:

$\theta = \tan^{-1} \left(\frac{2}{-3} \right)$

Since $-3 + 2i$ is in the second quadrant, we adjust $\theta$:

$\theta = \pi + \tan^{-1} \left(\frac{2}{-3} \right)$

Approximating:

$\tan^{-1}(-2/3) \approx -0.588$

$\theta \approx \pi - 0.588 \approx 2.553 \text{ radians}$

Thus, the polar form is:

$\sqrt{13} e^{i 2.553}$

Step 2: Use De Moivre’s Theorem

De Moivre’s Theorem states:

$(r e^{i\theta})^n = r^n e^{i n\theta}$

Applying $n = 5$:

$(\sqrt{13} e^{i 2.553})^5 = (\sqrt{13})^5 e^{i (5 \times 2.553)}$

Computing powers:

$(\sqrt{13})^5 = 13^{5/2} = 13^{2.5} = 13^2 \cdot 13^{0.5} = 169 \sqrt{13}$

Computing angle:

$5 \times 2.553 = 12.765$

Since $12.765$ radians is more than $2\pi$, we reduce it:

$12.765 - 2(3.1416) = 12.765 - 6.2832 = 6.4818$

$6.4818 - 2(3.1416) = 6.4818 - 6.2832 = 0.1986 \text{ radians}$

Step 3: Convert Back to Rectangular Form

$e^{i 0.1986} = \cos(0.1986) + i \sin(0.1986)$

Approximating:

$\cos(0.1986) \approx 0.980$ $\sin(0.1986) \approx 0.198$

Multiplying by $169 \sqrt{13}$:

$\text{Real part} = 169 \sqrt{13} \times 0.980 \approx 169 \times 3.605 \times 0.980$

$\approx 597.3$

$\text{Imaginary part} = 169 \sqrt{13} \times 0.198 \approx 169 \times 3.605 \times 0.198$

$\approx 120.7 i$

Final Answer:

$(-3+2i)^5 \approx 597.3 + 120.7 i$

Would you like a breakdown of any step?

Related Questions:

1. How do you convert a complex number to polar form?
2. What is De Moivre’s Theorem, and why is it useful?
3. How do you simplify large exponentiations of complex numbers?
4. What are the applications of complex numbers in engineering?
5. How do you find the principal argument of a complex number?

Tip:

Always check which quadrant your complex number is in when calculating the argument $\theta$.