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Exercise 4.1 · Q14

Q.Express the following expression in the form of a+iba + ib: (3+i5)(3−i5)(3+2 i)−(3−i2)\dfrac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - i\sqrt{2})}

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The numerator simplifies to a real number using the difference of squares, and the denominator simplifies to a pure imaginary number. Dividing gives a pure imaginary result: 0−722i0 - \frac{7\sqrt{2}}{2}i.

We start with the expression:

(3+i5)(3−i5)(3+2 i)−(3−i2)\frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - i\sqrt{2})}

The key is to simplify the numerator and denominator separately before attempting division. Complex numbers often look messy, but many problems are designed so that conjugates and simple arithmetic collapse them into clean forms.


1. Simplify the numerator

The numerator is a product of two conjugates: (3+i5)(3−i5)(3 + i\sqrt{5})(3 - i\sqrt{5}).

Recall the identity: (a+ib)(a−ib)=a2+b2(a + ib)(a - ib) = a^2 + b^2 (since i2=−1i^2 = -1). Here a=3a = 3 and b=5b = \sqrt{5}.

So:

(3+i5)(3−i5)=32+(5)2=9+5=14(3 + i\sqrt{5})(3 - i\sqrt{5}) = 3^2 + (\sqrt{5})^2 = 9 + 5 = 14

The numerator is a purely real number, 1414.

Tip

Whenever you see a product of conjugates, immediately think a2+b2a^2 + b^2 — it saves time and avoids sign errors.


2. Simplify the denominator

The denominator is:

(3+2 i)−(3−i2)(\sqrt{3} + \sqrt{2}\,i) - (\sqrt{3} - i\sqrt{2})

Distribute the minus sign carefully:

=3+2 i−3+i2= \sqrt{3} + \sqrt{2}\,i - \sqrt{3} + i\sqrt{2}

The 3\sqrt{3} terms cancel: 3−3=0\sqrt{3} - \sqrt{3} = 0.

Now combine the imaginary terms: 2 i+i2\sqrt{2}\,i + i\sqrt{2} — these are the same quantity, so:

=22 i= 2\sqrt{2}\,i

The denominator is a pure imaginary number.

Watch out

A common mistake is to forget that i2i\sqrt{2} and 2 i\sqrt{2}\,i are the same thing, or to mishandle the subtraction of the second bracket. Write it out term by term if needed.


3. Perform the division

We now have:

1422 i\frac{14}{2\sqrt{2}\,i} …

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