AHL 1.12—Complex numbers introduction

A complex number is written as $z=a+bi$, where $a$ is the real part and $b$ is the imaginary part. Note that both $a$ and $b$ are real numbers.

For $z=-7+4i$, comparing with $a+bi$ gives $a=-7$ and $b=4$.

Therefore $\text{Re}(z)=-7$ and $\text{Im}(z)=4$. The imaginary part is the coefficient of $i$, not the term $4i$ itself.

We need the real and imaginary parts of $z=-7+4i$.

Option B gives $\text{Im}(z)=4i$. The imaginary part is a real number (the coefficient of $i$), not a complex expression — so this is incorrect.

Option C has the same error for the imaginary part: $\text{Im}(z)=4i$ should be $4$.

Option D swaps the roles: $\text{Re}(z)=4$ and $\text{Im}(z)=-7$. This reverses the real and imaginary components.

Only option A correctly identifies $\text{Re}(z)=-7$ and $\text{Im}(z)=4$.

Powers of $i$ repeat with period 4: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$.

$38=4\times 9+2$, so the remainder is $2$.

$i^{38}=i^2=-1$.

We must find which value in $\{-1,1,i,-i\}$ equals $i^{38}$.

$i^n=i$ or $-i$ only when the remainder after dividing by 4 is 1 or 3 respectively. Since $38=4(9)+2$, the remainder is 2, so neither $i$ nor $-i$ is correct.

$i^n=1$ only when the remainder is 0 (i.e. $n$ divisible by 4). Since $38$ is not divisible by 4, option $1$ is incorrect.

Remainder $2$ gives $i^2=-1$, confirming the answer is $-1$.