Square Root of a Complex Number

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Root of minus one is not a real number, so we brought in iota, right? Now my question: what is the square root of iota itself? Do we need one more new number? Today we will find the square roots of any complex number, with one formula and one topper's trick. And listen, in JEE these come inside quadratic equations, and in tricky option questions. So, let's begin.

0:31Two square roots, always

So, first a basic question. Take a complex number z, not equal to zero; the crossed equals sign means not equal. How many square roots does it have? Exactly two, yes! Every non-zero complex number has exactly two square roots, negatives of each other. Find one, put a minus sign, and you have the other. That's it! And one difference from real numbers. There, root a needs a greater than or equal to zero. For complex numbers, there is no such condition. Okay?

1:05Why it works: building the formula

Alright, now let's build the formula together. Let z be a plus iota b, and suppose its square root is x plus iota y, with x and y real. Square both sides: x square minus y square, plus two iota x y, equals a plus iota b. The double arrow means implies. Now compare the real parts, and the imaginary parts. x square minus y square equals a, and two x y equals b. Simple. But solving these two directly is messy, right? So think: can we get x square plus y square some other way? Hint: use the modulus. Yes, exactly! Take mod on both sides; mod is the two vertical bars. Mod of x plus iota y, whole square, equals mod of a plus iota b. The left side is x square plus y square, and the right side is just mod z. Beautiful, isn't it? Now add and subtract these two equations. x square equals mod z plus a, over two. And y square equals mod z minus a, over two. Clear? But wait. From x square, x can be plus or minus, and from y square, y can be plus or minus too. So which pairs are allowed? Which equation decides? Two x y equals b, exactly! If b is positive, x and y have the same sign. If b is negative, they have opposite signs. So we get exactly two roots, not four.

2:40The square root formula (b≠0)

So, put it all together. Here is the formula, for b not equal to zero. Root of a plus iota b equals plus or minus, big bracket, root of mod z plus a, over two, plus iota times b over mod b, times root of mod z minus a, over two. And mod z is root of a square plus b square. This b over mod b is just the sign of b: it carries our sign rule. Got it?

3:04Topper's trick: spot the perfect square

Now let me share a topper's trick. The formula always works, but often there is a faster way: look for a perfect square directly. Remember, x plus iota y, whole square, is x square minus y square, plus two iota x y. Take root of minus five plus twelve iota. We need x square minus y square equal to minus five, and two x y equal to twelve, so x y equals six. Now guess: which simple numbers work? Two and three, correct! Four minus nine is minus five, and two into three is six; the dot means times. Both conditions are satisfied. So root of minus five plus twelve iota is plus or minus, two plus three iota. One line, no formula!

3:53The trap: √z has two values

Now, a trap. For real numbers, the root sign gives only the positive root, remember? So root of minus five plus twelve iota is just two plus three iota. That one looks positive, right? No, no, careful! For a complex z, root z has two values, and there is no positive root to prefer. Minus two minus three iota is just as much a square root as two plus three iota. So in JEE, if the root of a complex number appears, both signs may be intended. Work with plus or minus, and then check the options. Clear?

4:27Example 1: square roots of 2i and (1+i)/(1-i)

Alright, Example one. Simplify root of two iota, and root of, one plus iota over one minus iota. And look, the second one is connected to our puzzle from the start. Pause the video, and try it. Done? Okay, let's check. Part one: the trick. One plus iota, whole square, is one, plus two iota, minus one: two iota. So root of two iota is plus or minus, one plus iota. Part two: simplify inside the root first. Multiply the top and the bottom by one plus iota. The top becomes one plus iota whole square, two iota, from part one. The bottom, one minus iota times one plus iota, is two. So the inside is just iota! Aha, so we need root of iota. Our puzzle! Use the formula: here a is zero and b is one. So first, what is mod iota? One, correct! So x square is one plus zero, over two, a half. And y square is one minus zero, over two, also a half. And b is positive, so x and y have the same sign. So root of iota is plus or minus, one plus iota, over root two. There you go, puzzle solved! No new number needed: just two ordinary complex numbers, okay?

5:52Example 2: solve a quadratic with complex coefficients

Now a harder one, JEE level. Solve: z square, minus three minus two iota times z, plus five minus five iota, equals zero. Pause the video and try it. Hint: the quadratic formula works for complex coefficients too. Done? Okay, let's check. Quadratic formula: z equals, three minus two iota, plus or minus root D, all over two. And the discriminant D is three minus two iota whole square, minus four times five minus five iota. Three minus two iota, whole square, is five minus twelve iota. Then minus twenty, plus twenty iota. So D equals minus fifteen plus eight iota. Now we need its square root. Use the formula. First, mod D, the modulus of D: root of fifteen square plus eight square. What is it? Seventeen, yes! So x square is seventeen minus fifteen, over two, which is one. y square is seventeen plus fifteen, over two, which is sixteen. And b, the eight, is positive, so same signs. So root D is plus or minus, one plus four iota. The plus or minus already sits in the formula, so we just put one plus four iota there, okay? With the plus sign: three minus two iota, plus one plus four iota, is four plus two iota; over two, that is two plus iota. With the minus sign: two minus six iota, over two, which is one minus three iota. Now, a classic JEE trap. Complex roots come in conjugate pairs, don't they? So if two plus iota is a root, then two minus iota must also be a root, right? No, no, careful! Two plus iota and one minus three iota are not conjugates at all! The conjugate pair rule needs real coefficients, and here they are not real. The quadratic formula works, but that rule does not. Got it?

7:52Quick revision: the formula, sign rule from b, two values

So, quick revision. Every non-zero complex number has two square roots, plus or minus: x square is mod z plus a, over two, and y square is mod z minus a, over two. Signs: b positive, same; b negative, opposite. Or just spot the perfect square. Root z has two values, and with complex coefficients, the roots need not be conjugates. Now go and try the practice problems on this page. All the best!