Maths Support: Complex arithmetic

Question 1

Express the following in the form $a+bi\;$ where $a$ and $b$ are real.

Do not include decimals in your answers, only fractions or integers. Also do not include brackets in your answers.

a)

$(\simplify[std]{{a}})(\simplify[std]{{b}})\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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b)

$(\simplify[std]{{a1}})^2\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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c)

$\simplify[std,!otherNumbers]{{a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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d)

$(\simplify[std]{{z1}}) (\simplify[std]{{z2}}) (\simplify[std]{{z3}})\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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a)
The formula for multiplying complex numbers is
\[\begin{eqnarray*}\simplify[]{Re((a + ib)(c + id))} &=& ac -bd \\ \simplify[]{Im((a + ib)(c + id))} &=& ad +bc \end{eqnarray*} \]

So we have:
\[\begin{eqnarray*}\simplify[]{Re({a}*{b})} &=& \simplify[]{{Re(a)}*{Re(b)} - {Im( a)}*{Im(b)} = {Re(a*b)}}\\ \simplify[]{Im({a}*{b})} &=& \simplify[]{{Re(a)}*{Im(b)} + {Im( a)}*{Re(b)} = {Im(a*b)}} \end{eqnarray*} \]
Hence the solution is :

\[(\simplify[std]{{a}})(\simplify[std]{{b}})=\var{a*b}\]
b)

This is calculated in a similar way once the expression is written as:

$(\simplify[std]{{a1}})^2= (\simplify[std]{{a1}}) (\simplify[std]{{a1}})$ then we find:

\[\begin{eqnarray*}(\simplify[std]{{a1}})^2&=& (\simplify[std]{{a1}}) (\simplify[std]{{a1}})\\ &=& \simplify[]{({Re(a1)}*{Re(a1)} - {Im(a1)}*{Im(a1)})+ ({Re(a1)}*{Im(a1)} + {Im(a1)}*{Re(a1)})i}\\ &=& \simplify[std]{{a1^2}} \end{eqnarray*} \]
c)
We know that $i^2=-1$ which gives $i^3=i^2i=-i$.

Hence:
\[ \begin{eqnarray*} \simplify[std,!otherNumbers]{{a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3}&=&\simplify[std]{{a3} + {b3} * i -{c3} -({d3} * i)}\\ &=&\simplify[std]{ {a3} -{c3} + ({b3} -{d3}) * i}\\ &=&\simplify[std]{{a3 -c3} + {b3 -d3} * i} \end{eqnarray*} \]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
and then to multiply the result of that calculation by the third set of parentheses.

So we obtain:
\[ \begin{eqnarray*} (\var{z1})(\var{z2})(\var{z3})&=&((\var{z1})(\var{z2}))(\var{z3})\\ &=&(\var{z1*z2})(\var{z3})\\ &=&\var{z1*z2*z3} \end{eqnarray*} \]

Question 2

Express the following in the form $a+bi$.

Input $a$ and $b$ as fractions or integers and not as decimals.

a)

$\displaystyle \simplify[std]{{c1}/{z1}}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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b)

$\displaystyle \simplify[std]{{c2}/{z2}}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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c)

$\displaystyle \simplify[std]{{z1}/{z3}}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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d)

$\displaystyle \simplify[std]{{z3}/{z2}}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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Advice

Division of two complex numbers can be performed by mutiplying both the numerator and denominator by the conjugate of the denominator.
Suppose that \[ z = \frac{a+bi}{c+di},\;\; c+di \neq 0\] then we have:
\[\begin{eqnarray*} z&=&\frac{a+bi}{c+di}\\ &=&\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\ &=&\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\ &=&\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i \end{eqnarray*} \]
Although this is a formula for the inverse, the best way to find these complex numbers is to remember to multiply top and bottom by the conjugate of the denominator.
(a)
\[\begin{eqnarray*}\simplify[std]{{c1}/{z1}} &=&\simplify[std]{({c1}*{conj(z1)})/({z1}*{conj(z1)})}\\ &=&\simplify[std]{{c1*conj(z1)}/{abs(z1)^2}}\\ &=& \simplify[std]{{c1*re(z1)}/{abs(z1)^2}-{c1*im(z1)}/{abs(z1)^2}*i} \end{eqnarray*} \]
(b)
\[\begin{eqnarray*}\simplify[std]{{c2}/{z2}} &=&\simplify[std]{({c2}*{conj(z2)})/({z2}*{conj(z2)})}\\ &=&\simplify[std]{{c2*conj(z2)}/{abs(z2)^2}}\\ &=& \simplify[std]{{c2*re(z2)}/{abs(z2)^2}-{c2*im(z2)}/{abs(z2)^2}*i} \end{eqnarray*} \]
(c)
\[\begin{eqnarray*}\simplify[std]{{z1}/{z3}} &=&\simplify[std]{({z1}*{conj(z3)})/({z3}*{conj(z3)})}\\ &=&\simplify[std]{{z1*conj(z3)}/{abs(z3)^2}}\\ &=& \simplify[std]{{re(z1*conj(z3))}/{abs(z3)^2}+{im(z1*conj(z3))}/{abs(z3)^2}*i} \end{eqnarray*} \]
(d)
\[\begin{eqnarray*}\simplify[std]{{z3}/{z2}} &=&\simplify[std]{({z3}*{conj(z2)})/({z2}*{conj(z2)})}\\ &=&\simplify[std]{{z3*conj(z2)}/{abs(z2)^2}}\\ &=& \simplify[std]{{re(z3*conj(z2))}/{abs(z2)^2}+{im(z3*conj(z2))}/{abs(z2)^2}*i} \end{eqnarray*} \]

Question 3

Find the following complex numbers in the form $a+bi\;$ where $a$ and $b$ are real.

Input all numbers as fractions or integers. Also do not include brackets in your answers.

a)

$\var{e6*i}(\simplify[std]{{a}})\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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b)

$(\simplify[std]{{a}})(\simplify[std]{{z4}})\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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c)

$\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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d)

$(\simplify[std]{{a}})(\simplify[std]{ {z1}})(\simplify[std]{ {z3}})\;=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

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Advice

The solution is given by:

$\simplify[std]{{e6*i}}(\simplify[std]{{a}})=\simplify[std]{{a*e6*i}}$

$\simplify[std]{{a}*{z4}={a*z4}}$

c)
\[ \begin{eqnarray*} \simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}&=&\simplify[std]{{a}*{a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3}}\\ &=&\simplify[std]{{a*(a3 + b3 * i + c3 * i ^ 2 + d3 * i ^ 3)}} \end{eqnarray*} \]
d)

This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,

and then to multiply the result of that calculation by the third set of parentheses.

So we obtain:
\[ \begin{eqnarray*} (\var{a})(\var{z1})(\var{z3})&=&((\var{a})(\var{z1}))(\var{z3})\\ &=&(\var{a*(z1)})(\var{z3})\\ &=&\var{a*(z1)*(z3)} \end{eqnarray*} \]

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Maths Support: Complex arithmetic

Question 1

Question 1

a)

b)

c)

d)

Advice

Question 2

Question 2

a)

b)

c)

d)

Advice

Question 3

Question 3

a)

b)

c)

d)

Advice