Complex Numbers - Complex Analysis

Card 0 of 104

Question

Evaluate: $\left| 9 - 7 i \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$ .

We apply this notion to get.

$\left| 9 - 7 i \right| = \sqrt{ 81 + 49 }$

$= \sqrt{130}$

Compare your answer with the correct one above

Question

Evaluate: $\left| 1 - 3 i \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| 1 - 3 i \right| = \sqrt{ 1 + 9 }$

$= \sqrt{ 10 }$

Compare your answer with the correct one above

Question

Evaluate:

$\left| 9 - 9 i \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| 9 - 9 i \right| = \sqrt{ 81 + 81 }$

$= \sqrt{ 162 }$

$= 9 \sqrt{2}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| 6 - 3 i \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| 6 - 3 i \right| = \sqrt{ 36 + 9 }$

$= \sqrt{ 45 }$

$= 3 \sqrt{5}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 9 + 2 i ) ( -9 + 4 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 9 + 2 i ) ( -9 + 4 i ) \right| = \sqrt{ 81 + 4 }\cdot \sqrt{ 81 + 16 }$

$= \sqrt{85} \cdot \sqrt{97}$

$= \sqrt{8245}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 7 + 6 i ) ( -8 + 6 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 7 + 6 i ) ( -8 + 6 i ) \right| = \sqrt{ 49 + 36 }\cdot \sqrt{ 64 + 36 }$

$= \sqrt{85} \cdot 10$

$= 10 \sqrt{85}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 9 + 7 i ) ( -5 + 6 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 9 + 7 i ) ( -5 + 6 i ) \right| = \sqrt{ 81 + 49 }\cdot \sqrt{ 25 + 36 }$

$= \sqrt{130} \cdot \sqrt{61}$

$= \sqrt{7930}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 5 + 6 i ) ( -8 + 5 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 5 + 6 i ) ( -8 + 5 i ) \right| = \sqrt{ 25 + 36 }\cdot \sqrt{ 64 + 25 }$

$= \sqrt{61} \cdot \sqrt{89}$

$= \sqrt{5429}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 10 + 3 i ) ( -5 + 4 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 10 + 3 i ) ( -5 + 4 i ) \right| = \sqrt{ 100 + 9 }\cdot \sqrt{ 25 + 16 }$

$= \sqrt{109} \cdot \sqrt{41}$

$= \sqrt{4469}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 9 + i ) ( -9 + 6 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 9 + i ) ( -9 + 6 i ) \right| = \sqrt{ 81 + 1 }\cdot \sqrt{ 81 + 36 }$

$= \sqrt{82} \cdot 3 \sqrt{13}$

$= 3 \sqrt{1066}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 8 + 5 i ) ( -5 + 5 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 8 + 5 i ) ( -5 + 5 i ) \right| = \sqrt{ 64 + 25 }\cdot \sqrt{ 25 + 25 }$

$= \sqrt{89} \cdot 5 \sqrt{2}$

$=5 \sqrt{178}$

Compare your answer with the correct one above

Question

Evaluate:

$\left| ( 8 + 5 i ) ( -10 + 3 i ) \right|$

Answer

The general formula to figure out the modulus is

$\left|a+bi\right|=\sqrt{a^2+b^2}$

We apply this to get

$\left| ( 8 + 5 i ) ( -10 + 3 i ) \right| = \sqrt{ 64 + 25 }\cdot \sqrt{ 100 + 9 }$

$= \sqrt{89} \cdot \sqrt{109}$

$= \sqrt{9701}$

Compare your answer with the correct one above

Question

Evaluate $(\sqrt{3} - i)^6$

Answer

Converting from rectangular to polar coordinates gives us

$r = \sqrt{x^2+y^2} = \sqrt{3+1} = 2$

$\theta = tan^{-1}(y/x) = tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$

So

$(\sqrt{3}-i)^6 = (2e^{-\frac{\pi}{6}i})^6 = 64e^{-\pi i} = -64$

Compare your answer with the correct one above

Question

Compute $(-16)^\frac{1}{4}$

Answer

Converting from Rectangular to Polar Coordinates

$(-16)^\frac{1}{4} = (16e^{i\pi + 2\pi k})^\frac{1}{4} = 2e^{i\frac{\pi}{4} + \frac{\pi}{2}k}$

Evaluating for

we get that

$\pm\sqrt{2}(1+i) \text{ and } \pm\sqrt{2}(1-i)$

Compare your answer with the correct one above

Question

What is the magnitude of the following complex number?

Answer

The magnitude of a complex number is defined as

$\vert a+bi\vert=\sqrt{a^2+b^2}.$

So the modulus of is

$\vert5+2i\vert=\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$ .

Compare your answer with the correct one above

Question

What is the magnitude of the following complex number?

Answer

The magnitude of a complex number is defined as

$\vert a+bi\vert=\sqrt{a^2+b^2}.$

So the modulus of is

$\vert3-i\vert=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$ .

Compare your answer with the correct one above

Question

What is the magnitude of the following complex number?

Answer

The magnitude of a complex number is defined as

$\vert a+bi\vert=\sqrt{a^2+b^2}.$

Because the complex number has no imaginary part, we can write it in the form . Then the modulus of is

$\vert8+0i\vert=\sqrt{8^2+0^2}=\sqrt{64+0}=\sqrt{64}=8$ .

Compare your answer with the correct one above

Question

What is the argument of the following complex number?

Answer

Note that the complex number lies in the first quadrant of the complex plane.

For a complex number , the argument of is defined as the real number $\theta$ such that

$\tan(\theta)=\frac{b}{a}$ ,

where $-\pi<\theta\leq\pi$ is in radians.

Then the argument of is

$\tan(\theta)=\frac{1}{1}=1$

$\theta=\frac{-3\pi}{4},\frac{\pi}{4}$ .

The angle $\frac{-3\pi}{4}$ lies in the third quadrant of the complex plane, but the angle $\frac{\pi}{4}$ lies in the first quadrant, as does . So $\theta=\frac{\pi}{4}$ .

Compare your answer with the correct one above

Question

What is the argument of the following complex number in radians, rounded to the nearest hundredth?

Answer

Note that the complex number lies in the fourth quadrant of the complex plane.

For a complex number , the argument of is defined as the real number $\theta$ such that

$\tan(\theta)=\frac{b}{a}$ ,

where $-\pi<\theta\leq\pi$ is in radians.

Then the argument of is

$\tan(\theta)=\frac{-3}{2}$

$\theta=-0.98,2.16$ .

The angle lies in the second quadrant of the complex plane, but the angle lies in the fourth quadrant, as does . So $\theta=-0.98$ .

Compare your answer with the correct one above

Question

Which of the following is equivalent to this expression?

$(\sqrt3+i)^5$

Answer

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

Any nonzero complex number can be written in the form $re^{\theta i}$ , where
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 , 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 .
$a=r\cos(\theta)=32\cos(\frac{5\pi}{6})=32(\frac{-\sqrt3}{2})=-16\sqrt3$
$b=r\sin(\theta)=32\sin(\frac{5\pi}{6})=32(\frac{1}{2})=16$

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

Compare your answer with the correct one above

Tap the card to reveal the answer