Question

1）Find the square of a complex number z=a+ib using only two real number multiplications plus as many additions and subtractions as you wish.

2）Multiply two complex numbers z1=a+ib and z2=c+id using only 3 real number multiplications.

3）Compute (a+ib)²(c+id)² using only 5 real number multiplications.

Answer 1

1.) Square of two complex numbers.

A complex number is a two-part number. It has a real part and an imaginary part. We tend to write it in the form,

a + bi, where i is the square root of negative one, i.e., (-1)^(1/2)

Meanwhile, the square of a number is the number times itself. This means that

(a + bi)^2 = (a +bi) * (a + bi)

We encountered something similar to this when we considered factors of quadratic equations. There is a systematic approach for expanding the product of two two-part factors. You may have encountered the acronym “FOIL”:

Multiply the two First terms
Multiply the two Outer terms
Multiply the two Inner terms
Multiply the two Last terms
Sum the four terms for the answer

Apply the same FOIL approach, with (a + bi) * (a + bi), getting

a^2 + abi + abi + (bi)^2

We can reorganize a bit. The middle two terms are the same, so we can list them once, but multiplied by two.

a^2 +2abi + (bi)^2

And now, we will look at that last term, and realize that the square of a product can be written as the product of the separate squares. (x * y)^2 = x^2 * y^2.

Let’s apply that rule:

a^2 + 2abi + ((b^2)*(i^2))

But “i” is the square root of -1. The square of the square root of a number is the number, itself. So (i^2) = (-1)^((1/2)*2) = (-1)^1 = (-1).

Let’s plug this in.

a^2 + 2abi + ((b^2)*(-1))

That last term is still ugly. We can commute the “times negative one) to the other side, and rewrite the whole term as a subtraction.

a^2 + 2abi - b^2

But looking at the expression, we don’t follow the format of a real part followed by an imaginary part. We have a real part, an imaginary part and another real part. Let’s re-group the real parts together.

a^2 - b^2 + 2abi

(7 + 3i)^2 = 7^2 - 3^2 + (2*7*3)i = 49 - 9 + 42i = 40 + 42i

2.) To multiply complex numbers:

Each part of the first complex number gets multiplied by
each part of the second complex number

Just use "FOIL", which stands for "Firsts, Outers, Inners, Lasts".
(a+bi)(c+di) = ac + adi + bci + bdi2

• Firsts: a × c
• Outers: a × di
• Inners: bi × c
• Lasts: bi × di

Example: (3 + 2i)(1 + 7i)
(3 + 2i)(1 + 7i)   = 3×1 + 3×7i + 2i×1+ 2i×7i
= 3 + 21i + 2i + 14i2
= 3 + 21i + 2i − 14 (because i2 = −1)
= −11 + 23i

3.) Compute (a+ib)²(c+id)^{2 :}

we know that ,

(a+ib)² = a2 + 2abi + b2i2 (1)

(c+id)² = c2 + 2cdi + c2i2 (2)

Now, from (1) & (2)   

(a+ib)²(c+id)² = ( a2 + 2abi + b2i2 ) ( c2 + 2cdi + c2i2 )

where,

${\begin{aligned}i^{0}&{}=1,\quad &i^{1}&{}=i,\quad &i^{2}&{}=-1,\quad &i^{3}&{}=-i,\\i^{4}&={}1,\quad &i^{5}&={}i,\quad &i^{6}&{}=-1,\quad &i^{7}&{}=-i,\end{aligned}}$

and you can evaluate it further by putting the values .

#Note: Hope the above content is useful for you and you learn some thing from this content .