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Algebra Tutoring

Complex Numbers

Till date, you have learnt any number, negative or positive, when squared can’t ever be a negative integer. Very true, it’s also practically proven that when you multiply a negative with another negative, it’s got to be positive. But till date, you haven’t come across “iota”. Yes, iota denoted by “i “, is an imaginary number which squares up to –1.

i² = -1

For argument’s sake we can say, since i=√-1,

Hence,   i²= (√-1)² = √(-1)² = √1 = 1 , but this is not correct and is not even necessary in this case as we already have two numbers, ±1 , which squares up to 1, so “iota” wouldn’t have been needed.
Also, if we see the exponential series of iota (i) follow a cyclic pattern.

i= √-1                    i5= (i4) (i1) = 1 (i1) = (i)
i²= -1                     i6= (i4) (i2) = 1(-1) = -1
i³= (-i)                   i7= (i4) (i3) = 1(-i) = (-i)
i4= 1                       i8= (i4) (i4) = 1(1) = 1

So, it’s evident that whenever you need to calculate any higher exponents of (i) it’s always easy and advisable that you convert the figure in the exponent into multiples of 4 and the solution will be either (i) or (i2) or (i3). In few cases it can (i0) too, i.e. 1, as anything raised to the power of zero is equals to 1 which also applies to (i). For example:

i400= i4×100+0= 1100 X i0= 1 X 1 = 1

This number, imaginary or iota as you may call it, helps you solve negative figures within square roots, but it also restricts you to use the traditional rules of squares and square roots. Hence, whenever you are solving a complex number numerical, solve the i-part first.

But what are complex numbers then?

Complex numbers are any number that has a real as well as imaginary part to it. For example, (a + bi) is a basic ‘standard’ complex number where “a” is the real part and “bi” is the imaginary part. We will now see mathematical operations on complex numbers.

(3+4i) + (4–3i) = (3+4) + (4i–3i) = (7+i)   [Note: you can only simplify the operation in these cases]

(6–7i) – (4–2i) = (6–4–7i+2i) = (2–5i)

(3–2i)(2+3i) = (6–4i–4i–6i²) = (6–8i–6(–1)) [since i²= -1] i.e.= (6–8i+6) = (12–8i)

A simple one would be (4/5i). Now as (i) can’t be kept in the denominator hence we will multiply both numerator and denominator by (i) i.e. (4i/5i²) = (4i/5(-1)) = (4i/-5) = -4/5(i)

Now let’s see a tough one:

6/(3+i)  Even if we multiply by (i) it won’t help.  Here we need “conjugates”. A conjugate is a complex number similar to another complex number with only the sign different in the middle i.e. (a+bi) is conjugate of (a–bi) & if we multiply both:

(a+bi)(a–bi) = a²-b²(i²) = a² – b²(-1) = a²+b²

6(3–i)/(3+i)(3–i) =(18–6i)/(3²+1²) = (18–6i)/10 = ( 18/10)– (6/10)(i)

This is all about complex numbers!