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Complex Analysis : Complex Numbers

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Example Questions

Example Question #21 : Complex Analysis

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

$\overline{z}=z$

Possible Answers:

The equation is never true.

The equation is always true.

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

The equation is only true if a=0 .

The equation is only true if b=0 .

Correct answer:

The equation is only true if b=0 .

Explanation:

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

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

So the equation is only true if b=0 .

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Example Question #21 : Complex Analysis

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

Possible Answers:

Not enough information is given.

Correct answer:

Explanation:

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

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

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Example Question #23 : Complex Analysis

Which of the following is equivalent to this expression?

$(\sqrt3+i)^5$

Possible Answers:

$5\sqrt3+5i$

32-i

$9\sqrt3+i$

$16+\sqrt3i$

None of these

Correct answer:

None of these

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #21 : Complex Analysis

Which of the following is equivalent to this expression?

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

Possible Answers:

$-\frac{1}{32}$

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

$4\sqrt2i$

None of these

Correct answer:

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #24 : Complex Analysis

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

then what is the value of ?

Possible Answers:

-2-2i

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

$-\sqrt2+\sqrt2i$

$2\sqrt2i$

None of these

Correct answer:

$-\sqrt2+\sqrt2i$

Explanation:

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

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

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

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

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

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

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Example Question #25 : Complex Analysis

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

$\overline{z}=-z$

Possible Answers:

The equation is only true if a=0 .

The equation is never true.

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

The equation is only true if b=0 .

The equation is only true if a=b .

Correct answer:

The equation is only true if a=0 .

Explanation:

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

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

So the equation is only true if a=0 .

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Complex Analysis : Complex Numbers

Study concepts, example questions & explanations for Complex Analysis

All Complex Analysis Resources

Example Questions

Example Question #21 : Complex Analysis

Example Question #21 : Complex Analysis

Example Question #23 : Complex Analysis

Example Question #21 : Complex Analysis

Example Question #24 : Complex Analysis

Example Question #25 : Complex Analysis

All Complex Analysis Resources

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