Question

If \( z(2 - i) = (3 + i) \), then \( z^{38} = \), (where \( z = x + iy \))

• A. \( - (2^{19})i \)
• B. \( 2^{19}i \)
• C. \( - (2^{19}) \)
• D. \( 2^{19} \)

Hint:

When raising a complex number to a high power, convert to polar form. Remember \(e^{i(9\pi + \pi/2)} = e^{i9\pi} \cdot e^{i\pi/2} = (-1) \cdot i = -i\).

Answer 1

The Correct Option is A

Step 1: Understand the task.
We are told $z(2-i) = 3+i$ and we must find $z^{38}$. First we solve for $z$, then raise it to the 38th power using polar form, which is the easy way for high powers.
Step 2: Solve for $z$.
Divide both sides by $2-i$ and multiply top and bottom by the conjugate $2+i$: \[ z = \frac{3+i}{2-i}\cdot\frac{2+i}{2+i} = \frac{(3+i)(2+i)}{4+1}. \]
Step 3: Expand the numerator.
$(3+i)(2+i) = 6 + 3i + 2i + i^2 = 6 + 5i - 1 = 5 + 5i$. So \[ z = \frac{5+5i}{5} = 1 + i. \]
Step 4: Write $z$ in polar form.
The size is $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$. The angle is $\frac{\pi}{4}$ because the real and imaginary parts are equal and positive. So $z = \sqrt{2}\,e^{i\pi/4}$.
Step 5: Raise to the 38th power.
Power both parts: \[ z^{38} = (\sqrt{2})^{38}\,e^{i\,38\pi/4} = 2^{19}\,e^{i\,19\pi/2}. \]
Step 6: Simplify the angle.
Write $\frac{19\pi}{2} = 9\pi + \frac{\pi}{2}$. Now $e^{i9\pi} = -1$ and $e^{i\pi/2} = i$, so the angle factor is $(-1)(i) = -i$. Hence \[ z^{38} = 2^{19}\cdot(-i) = -(2^{19})\,i. \] \[ \boxed{z^{38} = -(2^{19})\,i} \]