Question

If \( z_1=4i^{40}-5i^{35}+6i^{17}+2,\ z_2=-1+i \), then \( |z_1+z_2|= \)

• A. \(13\)
• B. \(5\)
• C. \(15\)
• D. \(12\)

Hint:

For powers of \(i\), divide the exponent by 4 and use the remainder to simplify.

Answer 1

The Correct Option is A

Concept: Powers of \(i\) repeat in a cycle of 4: \[ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1 \]

Step 1: Simplify \(i^{40}\). \[ 40 \div 4 \Rightarrow \text{remainder }0 \] So, \[ i^{40}=1 \]

Step 2: Simplify \(i^{35}\). \[ 35 \div 4 \Rightarrow \text{remainder }3 \] So, \[ i^{35}=i^3=-i \]

Step 3: Simplify \(i^{17}\). \[ 17 \div 4 \Rightarrow \text{remainder }1 \] So, \[ i^{17}=i \]

Step 4: Substitute in \(z_1\). \[ z_1=4(1)-5(-i)+6(i)+2 \] \[ z_1=6+11i \]

Step 5: Now add \(z_2=-1+i\). \[ z_1+z_2=(6+11i)+(-1+i) \] \[ z_1+z_2=5+12i \]

Step 6: Find modulus. \[ |5+12i|=\sqrt{5^2+12^2} \] \[ =\sqrt{25+144} \] \[ =\sqrt{169}=13 \] Therefore, \[ \boxed{13} \]