Question:
By simplifying
\[
i^{18}-3i^7+i^2(1+i^4)(i)^{22}
\]
we get
By simplifying
\[
i^{18}-3i^7+i^2(1+i^4)(i)^{22}
\]
we get
Show Hint
The powers of \(i\) repeat in a cycle of \(4\):
\[
i,\ -1,\ -i,\ 1.
\]
So, always divide the exponent by \(4\) and use the remainder to simplify powers of \(i\).
Updated On: Jun 22, 2026
- \(-1+3i\)
- \(1-3i\)
- \(1+3i\)
- \(-1-3i\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Use powers of \(i\).
We know that \[ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1 \] The powers of \(i\) repeat after every \(4\) powers.
Step 2: Simplify \(i^{18}\).
\[ i^{18}=i^{16}i^2 \] Since \[ i^{16}=(i^4)^4=1 \] therefore, \[ i^{18}=i^2=-1 \]
Step 3: Simplify \(-3i^7\).
\[ i^7=i^4i^3 \] Since \[ i^4=1 \] and \[ i^3=-i \] we get \[ i^7=-i \] Thus, \[ -3i^7=-3(-i)=3i \]
Step 4: Simplify \(i^2(1+i^4)(i)^{22}\).
Since \[ i^2=-1 \] and \[ i^4=1 \] we get \[ 1+i^4=1+1=2 \] Now, \[ i^{22}=i^{20}i^2 \] Since \[ i^{20}=(i^4)^5=1 \] therefore, \[ i^{22}=i^2=-1 \] So, \[ i^2(1+i^4)i^{22}=(-1)(2)(-1)=2 \]
Step 5: Add all simplified terms.
Now, \[ i^{18}-3i^7+i^2(1+i^4)i^{22} = -1+3i+2 \] \[ =1+3i \]
Step 6: Final conclusion.
Therefore, \[ \boxed{1+3i} \]
We know that \[ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1 \] The powers of \(i\) repeat after every \(4\) powers.
Step 2: Simplify \(i^{18}\).
\[ i^{18}=i^{16}i^2 \] Since \[ i^{16}=(i^4)^4=1 \] therefore, \[ i^{18}=i^2=-1 \]
Step 3: Simplify \(-3i^7\).
\[ i^7=i^4i^3 \] Since \[ i^4=1 \] and \[ i^3=-i \] we get \[ i^7=-i \] Thus, \[ -3i^7=-3(-i)=3i \]
Step 4: Simplify \(i^2(1+i^4)(i)^{22}\).
Since \[ i^2=-1 \] and \[ i^4=1 \] we get \[ 1+i^4=1+1=2 \] Now, \[ i^{22}=i^{20}i^2 \] Since \[ i^{20}=(i^4)^5=1 \] therefore, \[ i^{22}=i^2=-1 \] So, \[ i^2(1+i^4)i^{22}=(-1)(2)(-1)=2 \]
Step 5: Add all simplified terms.
Now, \[ i^{18}-3i^7+i^2(1+i^4)i^{22} = -1+3i+2 \] \[ =1+3i \]
Step 6: Final conclusion.
Therefore, \[ \boxed{1+3i} \]
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