Divide complex numbers Algebra II

Here we have:

• z_1=3+4i
• z_2=2-i

Therefore:

z_1\div z_2=\dfrac{3+4i}{2-i}

Multiply top and bottom by the conjugate of 2-i :

\dfrac{3+4i}{2-i}=\dfrac{\left(3+4i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}

FOIL gives:

\dfrac{3+4i}{2-i}=\dfrac{6+8i+3i+4i^2}{\left(2-i\right)\left(2+i\right)}

Since i^2=-1, we have:

\dfrac{3+4i}{2-i}=\dfrac{6-4+11i}{\left(2-i\right)\left(2+i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{3+4i}{2-i}=\dfrac{2+11i}{4+1}

Here we have:

• z_1=2+3i
• z_2=1-2i

Therefore:

z_1\div z_2=\dfrac{2+3i}{1-2i}

Multiply top and bottom by the conjugate of 1-2i :

\dfrac{2+3i}{1-2i}=\dfrac{\left(2+3i\right)\left(1+2i\right)}{\left(1-2i\right)\left(1+2i\right)}

FOIL gives:

\dfrac{2+3i}{1-2i}=\dfrac{2+4i+3i+6i^2}{\left(1-2i\right)\left(1+2i\right)}

Since i^2=-1, we have:

\dfrac{2+3i}{1-2i}=\dfrac{2+4i+3i-6}{\left(1-2i\right)\left(1+2i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{2+3i}{1-2i}=\dfrac{-4+7i}{1^2+2^2}=\dfrac{-4+7i}{5}

Here we have:

• z_1=2-4i
• z_2=i

Therefore:

z_1\div z_2=\dfrac{2-4i}{i}

Multiply top and bottom by the conjugate of i :

\dfrac{2-4i}{i}=\dfrac{\left(2-4i\right)\left(-i\right)}{i\left(-i\right)}

FOIL gives:

\dfrac{2-4i}{i}=\dfrac{\left(2-4i\right)\left(-i\right)}{i\left(-i\right)}=\dfrac{-2i+4i^2}{-i^2}

Since i^2=-1, we have:

\dfrac{2-4i}{i}=\dfrac{-2i-4}{1}

\dfrac{2-4i}{i}=-4-2i

Here we have:

• z_1=1+2i
• z_2=1-i

Therefore:

z_1\div z_2=\dfrac{1+2i}{1-i}

Multiply top and bottom by the conjugate of 1-i :

\dfrac{1+2i}{1-i}=\dfrac{\left(1+2i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}

FOIL gives:

\dfrac{1+2i}{1-i}=\dfrac{1+2i+i+2i^2}{\left(1-i\right)\left(1+i\right)}

Since i^2=-1, we have:

\dfrac{1+2i}{1-i}=\dfrac{1+2i+i-2}{\left(1-i\right)\left(1+i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{1+2i}{1-i}=\dfrac{-1+3i}{1^2+1^2}=\dfrac{-1+3i}{2}

Here we have:

• z_1=5-6i
• z_2=2-3i

Therefore:

z_1\div z_2=\dfrac{5+6i}{2-3i}

Multiply top and bottom by the conjugate of 2-3i :

\dfrac{5+6i}{2-3i}=\dfrac{\left(5+6i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}

FOIL gives:

\dfrac{5+6i}{2-3i}=\dfrac{10+15i+12i+18i^2}{\left(2-3i\right)\left(2+3i\right)}

Since i^2=-1, we have:

\dfrac{5+6i}{2-3i}=\dfrac{10+27i-18}{\left(2-3i\right)\left(2+3i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{5+6i}{2-3i}=\dfrac{-8+27i}{2^2+3^2}=\dfrac{-8+27i}{13}

Here we have:

• z_1=7-2i
• z_2=5-2i

Therefore:

z_1\div z_2=\dfrac{7-2i}{5-2i}

Multiply top and bottom by the conjugate of 5-2i :

\dfrac{7-2i}{5-2i}=\dfrac{\left(7-2i\right)\left(5+2i\right)}{\left(5-2i\right)\left(5+2i\right)}

FOIL gives:

\dfrac{7-2i}{5-2i}=\dfrac{35+14i-10i-4i^2}{\left(5-2i\right)\left(5+2i\right)}

Since i^2=-1, we have:

\dfrac{7-2i}{5-2i}=\dfrac{35+4i+4}{\left(5-2i\right)\left(5+2i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{7-2i}{5-2i}=\dfrac{39+4i}{5^2+2^2}\dfrac{39+4i}{29}

Here we have:

• z_1=1+i
• z_2=1-i

Therefore:

z_1\div z_2=\dfrac{1+i}{1-i}

Multiply top and bottom by the conjugate of 1-i :

\dfrac{1+i}{1-i}=\dfrac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}

FOIL gives:

\dfrac{1+i}{1-i}=\dfrac{1+i+i+i^2}{\left(1-i\right)\left(1+i\right)}

Since i^2=-1, we have:

\dfrac{1+i}{1-i}=\dfrac{1+2i-1}{\left(1-i\right)\left(1+i\right)}

Furthermore, we know that \left(a+ib\right)\left(a-ib\right)=a^2+b^2. Therefore:

\dfrac{1+i}{1-i}=\dfrac{2i}{1^2+1^2}=i