10.8: Complex numbers
- Page ID
- 45141
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)When mathematics was first used, the primary purpose was for counting. Thus, they did not originally use negative numbers, zero, fractions, or irrational numbers. However, the ancient Egyptians quickly developed the need for “a part” and so they made up a new type of number, the ratio or fraction. The Ancient Greeks did not believe in irrational numbers (people were killed for believing otherwise). The Mayans of Central America later made up the number zero when they found use for it as a placeholder. Ancient Chinese Mathematicians made up negative numbers when they found use for them.
When working with radicals, we often work with radicands which are greater than or equal to zero. What about the case when the radicand is negative, especially with even roots? Previously, we said numbers like \(\sqrt{-4}\) were not real numbers, but what kind of number is it? In this event, we call numbers that contain square roots of negative numbers complex numbers. Before we get to the complex number, we discuss the imaginary unit.
Imaginary Unit
The imaginary unit, denoted by \(i\), is the number whose square is \(−1\), i.e., \[i^2=-1\text{ or }i=\sqrt{-1}\nonumber\]
Simplify \(\sqrt{-16}\) using the imaginary unit.
Solution
\[\begin{array}{rl}\sqrt{-16}&\text{Consider the negative as a factor of }-1 \\ \sqrt{-1\cdot 16}&\text{Apply the product property of square roots} \\ \sqrt{-1}\cdot\sqrt{16}&\text{Evaluate and rewrite }\sqrt{-1}\text{ as }i \\ 4i&\sqrt{-16}\text{ using the imaginary unit}\end{array}\nonumber\]
Simplify \(\sqrt{-24}\) using the imaginary unit.
Solution
For this example, we use techniques from simplifying radicals in addition to rewriting the radical with the imaginary unit.
\[\begin{array}{rl}\sqrt{-24}&\text{Consider the negative as a factor of }-1 \\ \sqrt{-1\cdot 24}&\text{Apply the product property of square roots} \\ \sqrt{-1}\cdot\sqrt{24}&\text{Simplify }\sqrt{24}\text{ and rewrite }\sqrt{-1}\text{ as }i \\ i\cdot\sqrt{4\cdot 6}&\text{Simplify the radical} \\ 2i\sqrt{6}&\sqrt{-24}\text{ using the imaginary unit}\end{array}\nonumber\]
Simplify \((3i)(7i)\).
Solution
\[\begin{array}{rl}(3i)(7i)&\text{Multiply} \\ 21i^2&\text{Apply the definition and rewrite }i^2\text{ as }-1 \\ 21(\color{blue}{-1}\color{black}{)}&\text{Multiply} \\ -21&\text{Result}\end{array}\nonumber\]
As a rule of thumb, we always rewrite \(\sqrt{-1}\) as \(i\), and \(i^2\) as \(-1\).
Complex Numbers
A complex number is a number of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(a\) is called the real part of \(a + bi\) and \(bi\) is called the imaginary part of \(a + bi\).
Express \(4+\sqrt{-64}\) as a complex number in the form \(a+bi\).
Solution
\[\begin{array}{rl}4+\sqrt{-64}&\text{Rewrite }\sqrt{-64}\text{ as factors }64\text{ and }-1 \\ 4+\sqrt{-1\cdot 64}&\text{Apply product property of square roots} \\ 4+\sqrt{-1}\cdot\sqrt{64}&\text{Simplify the radicals} \\ 4+8i&\text{Complex number}\end{array}\nonumber\]
Here, \(4\) is the real part and \(8i\) is the imaginary part. Together, they make a complex number.
Express \(7-\sqrt{-18}\) as a complex number in the form \(a+bi\).
Solution
\[\begin{array}{rl}7-\sqrt{-18}&\text{Rewrite }\sqrt{-18}\text{ as factors }18\text{ and }-1 \\ 7-\sqrt{-1\cdot 18}&\text{Apply product property of square roots} \\ 7-\sqrt{-1}\cdot\sqrt{18}&\text{Simplify }\sqrt{18}\text{ and rewrite }\sqrt{-1}\text{ as }i \\ 7-i\cdot\sqrt{9\cdot 2}&\text{Simplify the radical} \\ 7-3i\sqrt{2}&\text{Complex number}\end{array}\nonumber\]
Here, \(7\) is the real part and \(−3i\sqrt{2}\) is the imaginary part. Together, they make a complex number.
Simplify Expressions with Complex Numbers
When simplifying expressions with complex numbers, it is important that we rewrite any radicals that contain \(\sqrt{-1}\) or \(i^2\), replacing them by \(i\) and \(−1\), respectively. Then we simplify.
Simplify \(\sqrt{-6}\cdot\sqrt{-3}\).
Solution
We rewrite each factor using the imaginary unit, then apply the operation.
\[\begin{array}{rl}\sqrt{-6}\cdot\sqrt{-3}&\text{Rewrite the radicals with }i \\ (i\sqrt{6})(i\sqrt{3})&\text{Multiply} \\ i^2\cdot\sqrt{18}&\text{Rewrite }i^2\text{ as }-1\text{ and simplify the }\sqrt{18} \\ -1\cdot\sqrt{9\cdot 2}&\text{Simplify the radical} \\ -1\cdot 3\sqrt{2}&\text{Simplify the }-1\cdot 3 \\ -3\sqrt{2}&\text{Product}\end{array}\nonumber\]
Notice, even though we started with imaginary units, our product didn’t contain any because of the \(i^2\) term. Recall, every time we see an \(i^2\), we rewrite it as \(−1\), which contains no \(i\).
Simplify \(\dfrac{-15-\sqrt{-200}}{20}\).
Solution
We rewrite each term using the imaginary unit as needed, then apply the operation.
\[\begin{array}{rl}\dfrac{-15-\sqrt{-200}}{20}&\text{Rewrite the radical with }i\text{ and as a product of factors} \\ \dfrac{-15-\sqrt{-1\cdot 100\cdot 2}}{20}&\text{Simplify the radical} \\ \dfrac{-15-10i\sqrt{2}}{20}&\text{Factor a GCF from the numerator} \\ \dfrac{5(-3-2i\sqrt{2})}{20}&\text{Reduce the fraction by a factor of }5 \\ \dfrac{\color{blue}{\cancel{5}}\color{black}{(}-3-2i\sqrt{2})}{\color{blue}{\cancelto{4}{20}}}&\color{black}{\text{Rewrite}} \\ \dfrac{-3-2i\sqrt{2}}{4}&\text{Quotient}\end{array}\nonumber\]
The answer above will suffice, but if we wanted to rewrite \(\dfrac{-3-2i\sqrt{2}}{4}\) as a standard complex number, then we would rewrite the answer as \[-\dfrac{3}{4}-\dfrac{\sqrt{2}}{2}i\nonumber\] where \(-\dfrac{3}{4}\) is the real part and \(-\dfrac{\sqrt{2}}{2}i\) is the imaginary part.
Simplify Expressions with Complex Numbers by Adding, Subtracting, & Multiplying
We apply arithmetic operations to complex numbers in a way very similar to the way we apply arithmetic operations with expressions that contain variables. We combine like terms, when necessary. In this case, like terms are those with real parts and those with imaginary parts.
Add: \((2+5i)+(4-7i)\)
Solution
We simplify by combining like terms: combine real parts and combine imaginary parts.
\[\begin{array}{rl}(2+5i)+(4-7i)&\text{Combine like terms} \\ \underset{\color{blue}{\text{real parts}}}{\color{black}{\underbrace{(2+4)}}}+\underset{\color{blue}{\text{imaginary}}}{\color{black}{\underbrace{(5i-7i)}}}&\text{Simplify} \\ 6-2i&\text{Simplified expression}\end{array}\nonumber\]
Subtract: \((4-8i)-(3-5i)\)
Solution
We simplify by combining like terms: combine real parts and combine imaginary parts, but, first, we distribute the subtraction to each term in the parenthesis after the subtraction sign.
\[\begin{array}{rl}(4-8i)\color{blue}{-}\color{black}{(}3-5i)&\text{Distribute the negative} \\ 4-8i\color{blue}{-3+5i}&\color{black}{\text{Combine like terms}} \\ \underset{\color{blue}{\text{real parts}}}{\color{black}{\underbrace{(4-3)}}} + \underset{\color{blue}{\text{imaginary}}}{\color{black}{\underbrace{(5i-8i)}}}&\text{Simplify} \\ 1-3i&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \((5i)-(3+8i)+(-4+7i)\)
Solution
We simplify by combining like terms: combine real parts and combine imaginary parts, but, first, we distribute the subtraction to each term in the parenthesis after the subtraction sign.
\[\begin{array}{rl}(5i)\color{blue}{-}\color{black}{(}3+8i)+(-4+7i)&\text{Distribute the negative} \\ 5i\color{blue}{-3-8i}\color{black}{-}4+7i&\text{Combine like terms} \\ \underset{\color{blue}{\text{real parts}}}{\color{black}{\underbrace{(-3-4)}}}+\underset{\color{blue}{\text{imaginary}}}{\color{black}{\underbrace{(5i-8i+7i)}}} &\text{Simplify} \\ -7+4i&\text{Simplified expression}\end{array}\nonumber\]
Multiplying with complex numbers is similar to multiplying with variables except we rewrite every \[\sqrt{-1}\quad\text{as}\quad i\quad\text{and}\quad i^2\quad\text{as}\quad -1\nonumber\]
Simplify: \(5i(3i-7)\)
Solution
We multiply as usual applying the same exponent rules.
\[\begin{array}{rl}\color{blue}{5i}\color{black}{(}3i-7)&\text{Distribute }5i \\ 15\color{blue}{i^2}\color{black}{-}35i&\text{Rewrite }\color{blue}{i^2=-1} \\ 15\color{blue}{(-1)}\color{black}{-}35i&\text{Simplify} \\ -15-35i&\text{Simplified expression}\end{array}\nonumber\]
Multiplying expressions where the factors take a binomial form, we an apply the method of FOIL, a method we discussed in the exponents and polynomial chapter.
Recall, the FOIL method:
\[\begin{array}{ll} \textbf{F}\text{irst}&\text{-Multiply the first terms in each parenthesis} \\ \textbf{O}\text{uter}&\text{-Multiply the outer terms in each parenthesis} \\ \textbf{I}\text{nner}&\text{-Multiply the inner terms in each parenthesis} \\ \textbf{L}\text{ast}&\text{-Multiply the last terms in each parenthesis}\end{array}\nonumber\]
Simplify: \((2-4i)(3+5i)\)
Solution
We multiply this expression using the method of FOIL.
\[\begin{array}{rl}(2-4i)(3+5i)&\text{FOIL} \\ 6+10i-12i-20\color{blue}{i^2}&\color{black}{\text{Rewrite }}\color{blue}{i^2=-1} \\ 6+10i-12i-20\color{blue}{(-1)}&\color{black}{\text{Simplify}} \\ 6+10i-12i+20&\text{Combine like terms} \\ 26-2i&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \((4-5i)^2\)
Solution
We multiply this expression using either the method of FOIL or the perfect square trinomial formula, where \((A − B)^2 = A^2 − 2AB + B^2\). Let’s use the perfect square trinomial formula.
\[\begin{array}{rl}(4-5i)^2&\text{Apply the perfect square trinomial formula} \\ (4)^2-2(4)(5i)+(5i)^2&\text{Simplify} \\ 16-40i+25\color{blue}{i^2}&\color{black}{\text{Rewrite }}\color{blue}{i^2=-1} \\ 16-40i+25\color{blue}{(-1)}&\color{black}{\text{Simplify}} \\ 16-40i-25&\text{Combine like terms} \\ -9-40i&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \((3i)(6i)(2-3i)\)
Solution
We multiply this expression as usual and with distribution.
\[\begin{array}{rl}(3i)(6i)(2-3i)&\text{Multiple first two monomials} \\ \color{blue}{18i^2}\color{black}{(}2-3i)&\text{Distribute }\color{blue}{18i^2} \\ 36i^2-54\color{blue}{i^3}&\color{black}{\text{Rewrite }}\color{blue}{i^3=i^2\cdot i} \\ 36\color{blue}{i^2}\color{black}{-}54\color{blue}{i^2}\color{black}{\cdot}i&\text{Rewrite }\color{blue}{i^2=-1} \\ 36\color{blue}{(-1)}\color{black}{-}54\color{blue}{(-1)}\color{black}{i}&\text{Simplify} \\ -36+54i&\text{Simplified expression}\end{array}\nonumber\]
Simplify Expressions with Complex Numbers by Applying the Conjugate
Dividing with complex numbers is interesting if we have an imaginary part in the denominator. What do we do with an \(i\) in the denominator? Let’s think about \(i\) and its representation: \(i = \sqrt{-1}\). If there is an \(i\) in the denominator, then there is a square root in the denominator. Hence, we have to rationalize the denominator, but now using complex numbers.
To rationalize the denominator with only an imaginary part in the denominator, multiply the numerator and denominator by \(i\), e.g, \[\dfrac{1}{i}\cdot\dfrac{i}{i}\nonumber\]
Simplify: \(\dfrac{7+3i}{-5i}\)
Solution
We see that there is a \(−5i\) in the denominator. We can multiply the numerator and denominator by \(i\) to rewrite the denominator without \(i\), i.e., without a square root.
\[\begin{array}{rl}\dfrac{7+3i}{-5i}&\text{Multiply numerator and denominator by }i \\ \dfrac{(7+3i)}{-5i}\cdot\dfrac{i}{i}&\text{Distribute }i\text{ in numerator} \\ \dfrac{7i+3\color{blue}{i^2}}{-5\color{blue}{i^2}}&\color{black}{\text{Rewrite }}i^2=-1 \\ \dfrac{7i+3\color{blue}{(-1)}}{-5\color{blue}{(-1)}}&\color{black}{\text{Simplify}} \\ \dfrac{7i-3}{5}&\text{Simplified expression}\end{array}\nonumber\]
There are times where the given denominator is not just the imaginary part. Often, in the denominator, we have a complex number. In order to rationalize these denominators, we use the conjugate.
We rationalize denominators with complex numbers of the type \(a ± bi\) by multiplying the numerator and denominator by their conjugates, e.g., \[\dfrac{1}{a+bi}\cdot\dfrac{a-bi}{a-bi}\nonumber\]
The conjugate for
- \(a+bi\) is \(a-bi\)
- \(a-bi\) is \(a+bi\)
Simplify: \(\dfrac{2-6i}{4+8i}\)
Solution
We see that there is a \(4+8i\) in the denominator. We can multiply the numerator and denominator by \(4 − 8i\) to rewrite the denominator without \(i\), i.e., without a square root.
\[\begin{array}{rl}\dfrac{2-6i}{4+8i}&\text{Multiply numerator and denominator by conjugate} \\ \dfrac{2-6i}{4+8i}\cdot\dfrac{4-8i}{4-8i}&\text{Multiply numerator and denominator} \\ \dfrac{8-16i-24i+48\color{blue}{i^2}}{16-64\color{blue}{i^2}}&\color{black}{\text{Rewrite }}\color{blue}{i^2=-1} \\ \dfrac{8-16i-24i+48\color{blue}{(-1)}}{16-64\color{blue}{(-1)}}&\color{black}{\text{Simplify}} \\ \dfrac{8-16i-24i-48}{16+64}&\text{Combine like terms} \\ \dfrac{-40-40i}{80}&\text{Factor out GCF from numerator} \\ \dfrac{40(-1-i)}{80}&\text{Reduce out GCF from numerator} \\ \dfrac{\color{blue}{\cancel{40}}\color{black}{(}-1-i)}{\color{blue}{\cancelto{2}{80}}}&\color{black}{\text{Simplify}} \\ \dfrac{-1-i}{2}&\text{Simplified expression}\end{array}\nonumber\]
Powers of \(i\)
Let's take a look at powers of \(i\):
\[\begin{array}{lll}i^1=i & i^5=i & i^9=i \\ i^2=-1 & i^6=-1 & i^{10}=-1 \\ i^3=-i & i^7=-i & i^{11}=-i \\ i^4=1 & i^8=1 & i^{12}=1\end{array}\nonumber\]
Notice, after every fourth power of \(i\), the cycle starts over where every power that is a multiple of four is \(1\). Hence, for any power of \(i\), we can simplify easily by rewriting the power of \(i\) as a product of \(i\) that is a multiple of four and \(i\) raised to a power of at most \(3\). Let’s look at an example.
Simplify: \(i^{35}\)
Solution
Notice the power is \(35\), which equals \(32\) plus \(3\). We can rewrite the power as a sum of \(32\) and \(3\), then the expression as a product.
\[\begin{array}{rl}i^{35}&\text{Rewrite the power as a sum with the largest multiple of four} \\ i^{32+3}&\text{Rewrite as a product using product rule of exponents} \\ i^{32}\cdot i^3&\text{Simplify} \\ 1\cdot -i&\text{Multiply} \\ -i&\text{Simplified expression}\end{array}\nonumber\]
To find where to split the power of \(i\), we could divide the power by four. Then use the remainder to evaluate the expression. For example, in Example 10.8.17 , we could divide \(35\) by \(4\): \[35\div 4=8\text{ R}\color{red}{3}\nonumber\]
Then use the remainder \(3\) to evaluate \(i^{35}\), i.e., \[i^{35}=i^{\color{red}{3}}\color{black}{=}-i\nonumber\]
Simplify: \(i^{73}\)
Solution
Using the note above, let’s take the power \(73\) and divide by \(4\): \[73\div 4=18\text{ R}\color{red}{1}\nonumber\]
We can use the remainder to rewrite \(i^{73}\) as \[i^{73}=i^{\color{red}{1}}\color{black}{=i}\nonumber\]
Hence \(i^{73}=i\).
We can use the remainder method or the method displayed in Example 10.8.17 .
Simplify: \(i^{124}\)
Solution
Using the remainder method, let’s take \(124\) and divide by \(4\): \[124\div 4=31\text{ R}\color{red}{0}\nonumber\]
We can use the remainder to rewrtie \(i^{124}\) as \[i^{124}=i^{\color{red}{0}}\color{black}{=1}\nonumber\]
Hence \(i^{124} = 1\). Notice, the power \(124\) is a multiple of four, and we know that any power of \(i\) that is a multiple of four is one from the cycle for powers of \(i\).
Complex Numbers Homework
Simplify.
\(\sqrt{-64}\)
\(\sqrt{-9}\)
\(\sqrt{-81}\)
\(\sqrt{-121}\)
\(\sqrt{-100}\)
\(\sqrt{-45}\)
\(\sqrt{-90}\)
\(\sqrt{-420}\)
\(\sqrt{-245}\)
\((6i)(-8i)\)
\((-5i)(8i)\)
\((-8i)(4i)\)
\((-3i)(-4i)\)
\((-9i)(-4i)\)
\((3i)(5i)\)
\((-7i)^2\)
\(\sqrt{-10}\cdot\sqrt{-2}\)
\(\sqrt{-12}\cdot\sqrt{-2}\)
\(-7-\sqrt{-16}\)
\(-3+\sqrt{-121}\)
\(2-\sqrt{-25}\)
\(4+\sqrt{-4}\)
\(\dfrac{3+\sqrt{-27}}{6}\)
\(\dfrac{8-\sqrt{-16}}{4}\)
\(\dfrac{-4-\sqrt{-8}}{-4}\)
\(\dfrac{6+\sqrt{-32}}{4}\)
\(\dfrac{25-\sqrt{-75}}{5}\)
\(\dfrac{-10+\sqrt{-250}}{5}\)
\(\dfrac{15+\sqrt{-108}}{6}\)
\(\dfrac{12+\sqrt{-192}}{8}\)
\(3-(-8+4i)\)
\((7i)-(3-2i)\)
\((-6i)-(3+7i)\)
\((3-3i)+(-7-8i)\)
\((i)-(2+3i)-6\)
\((6+5i)^2\)
\((-7-4i)(-8+6i)\)
\((-4+5i)(2-7i)\)
\((-8-6i)(-4+2i)\)
\((1+5i)(2+i)\)
\(\dfrac{-9+5i}{i}\)
\(\dfrac{-10-9i}{6i}\)
\(\dfrac{-3-6i}{4i}\)
\(\dfrac{10-i}{-i}\)
\(\dfrac{4i}{-10+i}\)
\(\dfrac{8}{7-6i}\)
\(\dfrac{7}{10-7i}\)
\(\dfrac{5i}{-6-i}\)
\((3i)-(7i)\)
\(5+(-6-6i)\)
\((-8i)-(7i)-(5-3i)\)
\((-4-i)+(1-5i)\)
\((5-4i)+(8-4i)\)
\((-i)(7i)(4-3i)\)
\((8i)(-2i)(-2-8i)\)
\((3i)(-3i)(4-4i)\)
\(-8(4-8i)-2(-2-6i)\)
\((-6i)(3-2i)-(7i)(4i)\)
\((-2+i)(3-5i)\)
\(\dfrac{-3+2i}{-3i}\)
\(\dfrac{-4+2i}{3i}\)
\(\dfrac{-5+9i}{9i}\)
\(\dfrac{10}{5i}\)
\(\dfrac{9i}{1-5i}\)
\(\dfrac{4}{4+6i}\)
\(\dfrac{9}{-8-6i}\)
\(\dfrac{8i}{6-7i}\)
\(i^{77}\)
\(i^{48}\)
\(i^{62}\)
\(i^{154}\)
\(i^{251}\)
\(i^{68}\)
\(i^{181}\)
\(i^{51}\)