Complex Analysis : Complex Numbers

Study concepts, example questions & explanations for Complex Analysis

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Example Questions

Example Question #21 : Complex Analysis

Given a complex number \displaystyle z=a+bi, under what conditions is the following equation true?

\displaystyle \overline{z}=z

Possible Answers:

The equation is always true.

The equation is only true if \displaystyle b=0.

The equation is never true.

The equation is only true if \displaystyle a=0.

The equation is only true if  \displaystyle \vert z\vert>1.

Correct answer:

The equation is only true if \displaystyle b=0.

Explanation:

\displaystyle \overline{z} denotes the conjugate of \displaystyle z and is defined as
.

 

Substituting this into the equation and simplifying yields:
\displaystyle \begin{align*} a-bi&=a+bi\\ a-bi-a-bi&=0\\ -2bi&=0\\ b&=0. \end{align*}

 

So the equation is only true if \displaystyle b=0.

Example Question #21 : Complex Analysis

What is the value of  \displaystyle \vert e^{i\theta}\vert, where \displaystyle -\pi< \theta\leq\pi is in radians?

Possible Answers:

Not enough information is given.

\displaystyle i

\displaystyle 0

\displaystyle e

\displaystyle 1

Correct answer:

\displaystyle 1

Explanation:

The magnitude of a complex number \displaystyle z=re^{i\theta} is defined as
\displaystyle \vert z\vert=r.

 

If \displaystyle z=e^{i\theta}, then \displaystyle r=1, so \displaystyle \vert e^{i\theta}\vert=1.

Example Question #23 : Complex Analysis

Which of the following is equivalent to this expression?

\displaystyle (\sqrt3+i)^5

Possible Answers:

\displaystyle 5\sqrt3+5i

\displaystyle 32-i

\displaystyle 9\sqrt3+i

\displaystyle 16+\sqrt3i

None of these

Correct answer:

None of these

Explanation:

Note that \displaystyle \sqrt3+i lies in the first quadrant of the complex plane.

 

Any nonzero complex number \displaystyle a+bi can be written in the form \displaystyle re^{\theta i}, where
\displaystyle r^2=a^2+b^2 and
\displaystyle \tan(\theta)=\frac{b}{a}.
(We stipulate that  \displaystyle -\pi< \theta\leq\pi is in radians.)

Conversely, a nonzero complex number \displaystyle re^{\theta i} can be written in the form \displaystyle a+bi, where
\displaystyle a=r\cos(\theta) and
\displaystyle b=r\sin(\theta).

 

We can convert \displaystyle \sqrt3+i by using the formulas above:
\displaystyle r^2=a^2+b^2=(\sqrt3)^2+1^2=3+1=4
\displaystyle r=\sqrt4=2,
and
\displaystyle \tan(\theta)=\frac{b}{a}=\frac{1}{\sqrt3}
\displaystyle \theta=\frac{-7\pi}{6},\frac{\pi}{6}

Since \displaystyle \frac{\pi}{6} lies in the first quadrant of the complex plane, as does \displaystyle \sqrt3+i\displaystyle \theta = \frac{\pi}{6}.

So \displaystyle \sqrt3+i=2e^{\frac{\pi}{6}i}.

 

We now substitute this into our original expression and expand.
\displaystyle \begin{align*} (3+i)^5&=(2e^{\frac{\pi}{6}i})^5\\ &=2^5e^{5\frac{\pi}{6}i}\\ &=32e^{\frac{5\pi}{6}i} \end{align*}.

 

Finally, we convert this number back to the form \displaystyle a+bi.
\displaystyle a=r\cos(\theta)=32\cos(\frac{5\pi}{6})=32(\frac{-\sqrt3}{2})=-16\sqrt3
\displaystyle b=r\sin(\theta)=32\sin(\frac{5\pi}{6})=32(\frac{1}{2})=16

 

So our final answer is \displaystyle -16\sqrt3+16i.

Example Question #21 : Complex Analysis

Which of the following is equivalent to this expression?

\displaystyle \left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)^{10}

Possible Answers:

\displaystyle -\frac{1}{32}

\displaystyle i

\displaystyle \frac{\sqrt2}{8}+\frac{\sqrt2}{8}i

\displaystyle 4\sqrt2i

None of these

Correct answer:

\displaystyle i

Explanation:

Note that \displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i lies in the first quadrant of the complex plane.

 

Any nonzero complex number \displaystyle a+bi can be written in the form \displaystyle re^{\theta i}, where
\displaystyle r^2=a^2+b^2 and
\displaystyle \tan(\theta)=\frac{b}{a}.
(We stipulate that  \displaystyle -\pi< \theta\leq\pi is in radians.)

Conversely, a nonzero complex number \displaystyle re^{\theta i} can be written in the form \displaystyle a+bi, where
\displaystyle a=r\cos(\theta) and
\displaystyle b=r\sin(\theta).

 

We can convert \displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i by using the formulas above:
\displaystyle r^2=a^2+b^2=\left(\frac{\sqrt2}{2}\right)^2+\left(\frac{\sqrt2}{2}\right)^2=\frac{1}{2}+\frac{1}{2}=1
\displaystyle r=\sqrt1=1,
and
\displaystyle \tan(\theta)=\frac{b}{a}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1
\displaystyle \theta=\frac{-3\pi}{4},\frac{\pi}{4}

Since \displaystyle \frac{\pi}{4} lies in the first quadrant of the complex plane, as does \displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\displaystyle \theta=\frac{\pi}{4}.

So \displaystyle \frac{\sqrt2}{2}+\frac{\sqrt2}{2}i=e^{\frac{\pi}{4}i}.

 

We now substitute this into our original expression and expand.
\displaystyle \begin{align*} \left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)^{10}&=(e^{\frac{\pi}{4}i})^{10}\\ &=e^{\frac{10\pi}{4}i} \end{align*}.
Because \displaystyle -\pi< \theta\leq\pi,  we substitute \displaystyle \frac{10\pi}{4} with the coterminal angle \displaystyle \frac{2\pi}{4}=\frac{\pi}{2}.
\displaystyle \left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)^{10}=e^{\frac{\pi}{2}i}.

 

Finally, we convert this number back to the form \displaystyle a+bi.
\displaystyle a=r\cos(\theta)=\cos(\frac{\pi}{2})=0
\displaystyle b=r\sin(\theta)=\sin(\frac{\pi}{2})=1

 

So our final answer is \displaystyle 0+1i=i.

Example Question #24 : Complex Analysis

If

\displaystyle (-\sqrt2+\sqrt2i)(\frac{i}{4})(z)=1,

then what is the value of \displaystyle z?

Possible Answers:

\displaystyle -2-2i

\displaystyle \frac{\sqrt2}{2}-\frac{\sqrt2}{2}i

\displaystyle -\sqrt2+\sqrt2i

\displaystyle 2\sqrt2i

None of these

Correct answer:

\displaystyle -\sqrt2+\sqrt2i

Explanation:

Note that the \displaystyle 1 on the right side of the equation can be written as \displaystyle e^{0i}.

 

Multiplying the first two terms on the left side yields
\displaystyle (-\sqrt2+\sqrt2i)(\frac{i}{4})=\frac{-\sqrt2}{4}-\frac{\sqrt2}{4}i.
Note that this number lies in the third quadrant of the complex plane.

 

We now convert \displaystyle -\frac{\sqrt2}{4}-\frac{\sqrt2}{4}i from the form \displaystyle a+bi to the form \displaystyle re^{i\theta}using the identities
\displaystyle r^2=a^2+b^2 and
\displaystyle \tan(\theta)=\frac{b}{a}.
\displaystyle r^2=a^2+b^2=\left(-\frac{\sqrt2}{4}\right)^2+\left(-\frac{\sqrt2}{4}\right)^2=\frac{2}{16}+\frac{2}{16}=\frac{4}{16}=\frac{1}{4}
\displaystyle r=\sqrt{\frac{1}{4}}=\frac{1}{2}
\displaystyle \tan(\theta)=\frac{b}{a}=\frac{-\frac{\sqrt2}{4}}{-\frac{\sqrt2}{4}}=1
\displaystyle \theta=-\frac{3\pi}{4},\frac{\pi}{4}.
Since \displaystyle -\frac{3\pi}{4} lies in the third quadrant of the complex plane, as does \displaystyle -\frac{\sqrt2}{4}-\frac{\sqrt2}{4}i\displaystyle \theta=-\frac{3\pi}{4}.
So our new form is \displaystyle \frac{1}{2}e^{-\frac{3\pi}{4}i}.

 

Our equation now reduces to
\displaystyle (\frac{1}{2}e^{-\frac{3\pi}{4}i})(z)=e^{0i}.
We solve for z by dividing.
\displaystyle \begin{align*} z&=\frac{e^{0i}}{\frac{1}{2}e^{-\frac{3\pi}{4}i}}\\ &=\frac{1}{\frac{1}{2}}e^{0i-\frac{-3\pi}{4}i}\\ &=2e^{\frac{3\pi}{4}i}. \end{align*}

 

Finally, we convert this to the form \displaystyle a+bi by using the identities
\displaystyle a=r\cos(\theta) and
\displaystyle b=r\sin(\theta).
\displaystyle a=r\cos(\theta)=2\cos(\frac{3\pi}{4})=2(-\frac{\sqrt2}{2})=-\sqrt2
\displaystyle b=r\sin(\theta)=2\sin(\frac{3\pi}{4})=2(\frac{\sqrt2}{2})=\sqrt2.

 

So our final answer is \displaystyle z=-\sqrt2+\sqrt2i.

Example Question #25 : Complex Analysis

Given a complex number \displaystyle z=a+bi, under what conditions is the following equation true?

\displaystyle \overline{z}=-z

Possible Answers:

The equation is only true if \displaystyle a=0.

The equation is never true.

The equation is only true if  \displaystyle \vert z\vert = \vert\overline{z}\vert.

The equation is only true if \displaystyle b=0.

The equation is only true if \displaystyle a=b.

Correct answer:

The equation is only true if \displaystyle a=0.

Explanation:

\displaystyle \overline{z} denotes the conjugate of \displaystyle z and is defined as
\displaystyle \ovelrine{z}=a-bi.

 

Substituting this into the equation and simplifying yields:
\displaystyle \begin{align*} a-bi&=-(a+bi)\\ a-bi&=-a-bi\\ a-bi+a+bi&=0\\ 2a&=0\\ a&=0 \end{align*}

 

So the equation is only true if \displaystyle a=0.

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