The Conjugate of a Complex Number

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take three plus four iota, and three minus four iota. Neither is real, right? Add them: six. Real! Multiply them: twenty five. Real again! Two non-real numbers, with a real sum and a real product? Strange, isn't it? These two are conjugates, and today we see why this always happens. It comes up often in JEE, so let's begin.

0:30The conjugate: flip the sign of the imaginary part

So, take z equals x plus iota y, with x and y real. Its conjugate is written as z with a bar on top: z bar. The arrow means implies: z bar equals x minus iota y. Only the imaginary part changes sign. Okay? Quick check. What is the conjugate of three plus four iota? Three minus four iota, yes, exactly! The bar over the whole number means its conjugate. Some students flip both signs and write minus three minus four iota. No, no! That is minus z. The real part, three, stays as it is. Clear? And one picture for later: z bar is the mirror image of z in the real axis. We will use this in the next chapter. Got it?

1:18Sum, difference and product with z bar

Now, quick properties. Two bars on top mean the conjugate taken twice: flip the sign, then flip it back. So z bar bar is just z, right? Add z and z bar: the iota y terms cancel, leaving two x, twice the real part of z. Subtract, and the x cancels: two iota y, that is two iota times the imaginary part; the short forms on screen say real part and imaginary part. Don't drop that iota! With me? Now the most important one: z times z bar. x plus iota y, times x minus iota y. Sum times difference, so x square minus y square, right? No, no, careful! Classic trap. It is x square, minus iota y whole square, and iota square is minus one. So we get x square plus y square. That is mod z square; the two vertical bars mean the modulus, mod z. So z z bar is real, and greater than or equal to zero. Remember this! And our puzzle is solved! Three plus four iota and three minus four iota are conjugates. Sum: twice the real part, six. Product: nine plus sixteen, twenty five. Real both times. Okay?

2:36The bar passes through everything

Alright, two complex numbers now, z one and z two; the small one and two are just labels. The conjugate of a sum or difference is the sum or difference of the conjugates; the plus or minus sign covers both. Same for a product. Got it? Same for a quotient: z one bar by z two bar, with z two not zero. And for a power: the conjugate of z to the power n is z bar to the power n. The bar passes through everything. So, for P of z, a polynomial with real coefficients, the conjugate of P of z is P of z bar. Why real coefficients? Because a real number is its own conjugate, so the coefficients stay the same. Clear?

3:18Topper's trick: the conjugate test

Now, when does z equal its own conjugate? Only when there is no imaginary part to flip. So z is purely real if and only if z bar equals z; that is what the double headed arrow means. And if the real part is zero, z is iota y, and z bar is minus iota y, that is minus z. So purely imaginary, or zero, means z bar equals minus z. Okay? Here is the topper's trick. To show an expression E is real, don't write x plus iota y. Just take its conjugate, E bar, and check that E bar equals E, okay? For purely imaginary, E bar should come out as minus E instead. Much faster! Let's try it on this expression, which comes up often: z one z two bar, plus z one bar z two. Is it real? Take its conjugate. Yes, it is real! The bar passes through: z one bar z two, plus z one z two bar. Same two terms, swapped. So E bar equals E, and E is real; in fact, twice the real part of z one z two bar. Perfect!

4:29Conjugate roots theorem

Now one more beautiful result. Take a polynomial equation with real coefficients, and a root a plus iota b. Which other number must also be a root? Its conjugate, a minus iota b, exactly! Here b is not zero, the crossed equals sign, so the root is not real. This is the conjugate roots theorem. The proof is one line. P of z is zero, so P of z bar equals the conjugate of P of z: zero bar, which is zero. So z bar is a root too. Non-real roots of real polynomials come in conjugate pairs.

5:06Example 1: conjugate of two complex fractions

Clear? Now, Example one. Find the conjugate. Part one: one over, three plus four iota. Part two: three minus two iota, times two plus three iota, over, one plus two iota, times two minus iota. Pause the video and try it first. Done? Okay, let's check. Part one: multiply top and bottom by three minus four iota. The bottom is our puzzle product, twenty five! So we get three minus four iota over twenty five, and its conjugate is three plus four iota over twenty five. Part two. The top: six, plus nine iota, minus four iota, minus six iota square. Iota square is minus one, so twelve plus five iota. The bottom: two, minus iota, plus four iota, minus two iota square: four plus three iota. Now divide: multiply by four minus three iota, and the bottom becomes twenty five. The top: forty eight, minus thirty six iota, plus twenty iota, plus fifteen. Sixty three minus sixteen iota, over twenty five. That's our answer, right? No, no, wait! That is the number itself, and the question asks for its conjugate. So flip the sign: sixty three plus sixteen iota, over twenty five. Always read the question to the last word. Got it?

6:34Example 2: (z - 1)/(z + 1) purely imaginary gives |z| = 1

Now a harder one. Prove: if z is not minus one, and z minus one, over z plus one, is purely imaginary, then mod z equals one. Pause the video and try it. Hint: use the conjugate test. Done? Okay, let's check. Putting z equals x plus iota y would be long and messy. Instead, call the fraction w. If w is purely imaginary, what is w bar? Minus w, correct! That one line, w bar equals minus w, replaces all the messy algebra. Now the bar passes through the fraction, and the conjugate of one is one. So z bar minus one, over z bar plus one, equals minus, z minus one over z plus one. With me? Cross multiply; z is not minus one, so neither denominator is zero. z bar minus one, times z plus one, equals minus, z minus one, times z bar plus one. Expand. Left: z z bar, plus z bar, minus z, minus one. Right: minus z z bar, minus z, plus z bar, plus one. Plus z bar and minus z sit on both sides. They cancel! So, z z bar minus one equals minus z z bar plus one, and two z z bar equals two. But z z bar is mod z square! So mod z square is one, and mod z is one. Proved, without any x and y!

8:09Quick revision: z bar, z z bar, conjugate test

So, quick revision. z bar: flip the sign of the imaginary part only, and z z bar is mod z square, always real. To show E is real, check that E bar equals E; for purely imaginary, E bar should be minus E, okay? And with real coefficients, non-real roots come in conjugate pairs. Now try the practice problems on this page. All the best!