AHL 1.12—Complex numbers – Cartesian form and Argand diag

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator. Here, the conjugate of $w=1-2i$ is $\bar{w}=1+2i$.

$\frac{3+4i}{1-2i}\cdot \frac{1+2i}{1+2i}=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}$

$(1-2i)(1+2i)=1^2+2^2=1+4=5$.

$(3+4i)(1+2i)=3+6i+4i+8i^2=3+10i+8(-1)=3+10i-8=-5+10i$.

$\frac{-5+10i}{5}=-1+2i$ So the answer is $-1+2i$.

We need $\frac{3+4i}{1-2i}$ in the form $a+bi$. The real part and imaginary part of the quotient must satisfy specific conditions.

The denominator has modulus squared $|1-2i{|}^2=1+4=5$. The numerator times conjugate of denominator gives $-5+10i$. Dividing by 5 gives $-1+2i$, so the real part is $-1$, not $+1$. This eliminates $1+2i$ and $1-2i$.

The imaginary part of the product $(3+4i)(1+2i)$ is $6+4=10$, which is positive. Dividing by 5 gives $+2i$. This eliminates $-1-2i$.

The only remaining option is $-1+2i$, which matches our calculation. The answer is $-1+2i$.

The powers of $i$ repeat with period 4: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, then the cycle repeats.

Divide 47 by 4: $47=4\times 11+3$. The remainder is $3$.

Since the remainder is 3, we have $i^{47}=i^3=-i$. The answer is $-\mathbf{i}$.

The powers of $i$ cycle: $i,-1,-i,1,i,-1,-i,1,\dots$ with period 4. We need to determine which position in the cycle 47 corresponds to.

$i^n=1$ only when $n$ is divisible by 4. Since $47=4\times 11+3$, 47 is not divisible by 4. Eliminate $1$.

$i^n=-1$ when the remainder on dividing $n$ by 4 is 2. Since $47÷4$ has remainder 3, not 2, eliminate $-1$.

$i^n=i$ when the remainder is 1. Since the remainder is 3, not 1, eliminate $i$.

Remainder 3 corresponds to $i^3=-i$. The answer is $-\mathbf{i}$.