Products and Quotients of Complex Numbers in Polar Form - Pre-Calculus

Card 0 of 96

Find the value of ,where the complex number is given by .

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We note that by FOILing.

We also know that:

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

Since we know that,

We have then:

Since we know that:

, we use a=2 ,b=i

We have then:

Compute the following sum:

. Remember is the complex number satisfying .

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Note that this is a geometric series.

Therefore we have:

Note that,

= and since we have .

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

this shows that the sum is 0.

Find the following product.

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

Therefore,

Compute the magnitude of .

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We have

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

We know that

Thus this gives us,

.

Let , . Find a simple form of $\frac{a}{b}$ .

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We remark first that:

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

and we know that :

.

This means that:

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

Evaluate:

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

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To evaluate this problem we need to FOIL the binomials.

Now recall that

Thus,

Find the product , if

.

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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

.

Simplify:

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The expression can be rewritten as:

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

The correct answer is:

Find the product of the two complex numbers

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

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The product is

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

What is ?

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Since ,

the problem becomes,

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

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

Write

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

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

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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}$i$

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

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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

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$

Divide:

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

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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.

Recall that

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

Divide:

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide:

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

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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:

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

Divide.

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide.

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

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

Now, distribute and simplify.

Recall that

Divide.

$\frac{-9+i}{1-i}$

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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{-9+i}{1-i}$\right)\left($\frac{1+i}{1+i}$\right)$

Now, distribute and simplify.

Recall that