Question

If $Z_{1}=4i^{40}-5i^{35}+6i^{17}+2$, $Z_{2}=-1+i$ where $i=\sqrt{-1}$, then $|Z_{1}+Z_{2}|=$

• A. 5
• B. 13
• C. 12
• D. 15

Hint:

Logic Tip: The triplet $(5, 12, 13)$ is a very common Pythagorean triple. Recognizing standard Pythagorean triples can save you time when calculating the modulus of a complex number or the magnitude of a vector.

Answer 1

The Correct Option is B

Concept:
The powers of the imaginary unit $i$ follow a cyclical pattern of 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. For any integer exponent $n$, $i^n = i^{n \pmod 4}$. After simplifying the complex numbers into the form $a + ib$, the modulus is calculated as $|a + ib| = \sqrt{a^2 + b^2}$.
Step 1: Simplify the powers of $i$ in $Z_1$.
Given $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$. Let's find the remainder of each exponent when divided by 4: $$i^{40} = (i^4)^{10} = (1)^{10} = 1$$ $$i^{35} = i^{32} \cdot i^3 = (i^4)^8 \cdot i^3 = 1 \cdot (-i) = -i$$ $$i^{17} = i^{16} \cdot i^1 = (i^4)^4 \cdot i = 1 \cdot i = i$$
Step 2: Evaluate $Z_1$ into standard form.
Substitute the simplified powers back into $Z_1$: $$Z_1 = 4(1) - 5(-i) + 6(i) + 2$$ $$Z_1 = 4 + 5i + 6i + 2$$ Combine real and imaginary parts: $$Z_1 = 6 + 11i$$
Step 3: Add $Z_1$ and $Z_2$.
Given $Z_2 = -1 + i$. $$Z_1 + Z_2 = (6 + 11i) + (-1 + i)$$ $$Z_1 + Z_2 = (6 - 1) + (11i + i)$$ $$Z_1 + Z_2 = 5 + 12i$$
Step 4: Calculate the modulus of the sum.
Using the modulus formula $|a + ib| = \sqrt{a^2 + b^2}$: $$|Z_1 + Z_2| = |5 + 12i|$$ $$|Z_1 + Z_2| = \sqrt{5^2 + 12^2}$$ $$|Z_1 + Z_2| = \sqrt{25 + 144}$$ $$|Z_1 + Z_2| = \sqrt{169} = 13$$