Polar Coordinates and Complex Numbers - Pre-Calculus

Card 0 of 588

Question

Find the value of ${}(1+i)^{100}$ ,where the complex number is given by ${}i^2=-1$ .

Answer

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

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Question

Compute the following sum:

${}1+i+i^2+i^3....+i^{2015}$ . Remember is the complex number satisfying ${}i^{2}=-1$ .

Answer

Note that this is a geometric series.

Therefore we have:

${}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$

Note that,

${}i^{2016}$ = ${}({i}^{2})^{1008}$ and since ${}i^2=-1$ we have ${}{-1}^{1008}=1$ .

$\frac{1-1}{1-i}=\frac{0}{1-i}=0$

this shows that the sum is 0.

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Question

Find the following product.

${}(1-i)^{n}(1+i)^n$

Answer

Note that by FOILing the two binomials we get the following:

${}(1+i)(1-i)=1-i+i-i^2=2$

Therefore,

${}(1-i)^{n}(1-i)^{n}=2^{n}$

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Question

Compute the magnitude of ${}(1+i)^{n}$ .

Answer

We have

${}|(1+i)|=\sqrt{2}$ .

We know that ${}|z^n|=|z|^n$

Thus this gives us,

${}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ .

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Question

Let $a=(1+i)^{n}$ , . Find a simple form of $\frac{a}{b}$ .

Answer

We remark first that:

$\frac{1+i}{1-i}=\frac{1+i}{1-i} \frac{1+i}{1+i}$ $=\frac{(1+i)^2}{2}$

and we know that :

.

This means that:

$\frac{1+i}{1-i}=\frac{2i}{2}=i$

$\left(\frac{1+i}{1-i}\right)^n=i^{n}$

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Question

Evaluate:

$(2+i)\cdot(1+4i)$

Answer

To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

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Question

Find the product , if

.

Answer

To find the product , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.

Using this method we get the following,

and because

.

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Question

Simplify: $i^4 \times i^3$

Answer

The expression $i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ , the value of .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is:

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Question

Find the product of the two complex numbers

$\small 3+7i$ and $\small 7+10i$

Answer

The product is

$\small (3+7i)(7+10i)=21+30i+49i+70i^2=21-70+79i=-49+79i$

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Question

Solve for (there may be more than one solution).

Answer

To solve for the roots, just set equal to zero and solve for z using the quadratic formula ( $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ) : $z^3-z^2+z\rightarrow z(z^2-z+1)$ and now setting both and equal to zero we end up with the answers and $z=\frac{1\pm 3i}{2}$

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Question

What is $\frac{(i^2)}{1-i}$ ?

Answer

Since ,

the problem becomes,

$\frac{-1}{1-i}$

$=-1+\frac{1}{1-i}$

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Question

Write

$\small \frac{1+2i}{1+i}$

in the form $\small a+bi$ for some real numbers $\small a$ and $\small b$ .

Answer

The correct answer is

$\small \frac{3}{2}+\frac{1}{2}i$

Using simple algebra and multiplying the expression by the complex conjugate of the denominator, we get:

$\small \small \frac{1+2i}{1+i}=\frac{1+2i}{1+i}\cdot\frac{1-i}{1-i}= (1-i)\frac{1+2i}{2}$

$\small =\frac{1-i+2i-2i^2}{2}=\frac{1+i-(-2)}{2}=\frac{3+i}{2}=\frac{3}{2}+\frac{1}{2}i$

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Question

Simplify $\frac{3+5i}{2-4i}$ into a number of the form .

Answer

We have

$\frac{3+5i}{2-4i}$

Multiply by the complex conjugate of the denominator.

The complex conjugate is the denominator with the sign changed:

$\frac{3+5i}{2-4i}\times \frac{2+4i}{2+4i}$

Multiply fractions

$\frac{(3+5i)(2+4i)}{(2-4i)(2+4i)}$

FOIL the numerator and denominator

$\frac{6+12i+10i+20i^2}{4+8i-8i-16i^2}$

Apply the rule of :

$\frac{6+12i+10i-20}{4+8i-8i+16}$

Simplify.

$\frac{-14+22i}{20}$

Simplify further using the addition of fractions rule, then factor the i out of the 2nd fraction.

$-\frac{7}{10}+\frac{11}{10}i$

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Question

Divide:

$\frac{4+5i}{5-i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)$

Now, distribute and simplify.

$\left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)=\frac{20+29i+5i^2}{25-i^2}$

Recall that

$\frac{20+29i+5(-1)}{25-(-1)}=\frac{15+29i}{26}=\frac{15}{26}+\frac{29i}{26}$

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Question

Convert $5(\cos \frac{ 8 \pi }{5} + i \sin \frac{ 8 \pi }{5} )$ to rectangular form

Answer

To convert, evaluate the trig ratios and then distribute the radius:

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Question

Divide:

$\frac{7+i}{2-i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)$

Now, distribute and simplify.

$\left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)=\frac{14+9i+i^2}{4-i^2}$

Recall that

$\frac{14+9i+i^2}{4-i^2}=\frac{14+9i+(-1)}{4-(-1)}=\frac{13+9i}{5}=\frac{13}{5}+\frac{9i}{5}$

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Question

Divide.

$\frac{2-9i}{3+i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)$

Now, distribute and simplify.

$\left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{6-29i+9i^2}{9-i^2}$

Recall that

$\frac{6-29i+9i^2}{9-i^2}=\frac{6-29i+9(-1)}{9-(-1)}=\frac{-3-29i}{10}=-\frac{3}{10}-\frac{29i}{10}$

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Question

Divide:

$\frac{5-i}{3+6i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )$

Now, distribute and simplify.

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )=\frac{15-30i-3i+6i^{2}}{9-6i^{2}}$

Recall that

$\frac{15-33i-6}{9+6}$

Then combine like terms:

$\frac{9-33i}{15}$

Then since each term is a multiple of you can simplify:

$\frac{3-11i}{5}$

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Question

Divide.

$\frac{4+3i}{6-i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)$

Now, distribute and simplify.

$\left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)=\frac{24+22i+3i^2}{36-i^2}$

Recall that

$\frac{24+22i+3i^2}{36-i^2}=\frac{24+22i+3(-1)}{36-(-1)}=\frac{21+22i}{37}$

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Question

Divide.

$\frac{5-2i}{3-8i}$

Answer

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be .

$\left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)$

Now, distribute and simplify.

$\left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)=\frac{15+34i-16i^2}{9-64i^2}$

Recall that

$\frac{15+34i-16i^2}{9-64i^2}=\frac{15+34i-16(-1)}{9-64(-1)}=\frac{31+34i}{73}=\frac{31}{73}+\frac{34i}{73}$

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