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Complex numbers
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Two methods:

1. **"multiply, then modulus"**: First expand the brackets to get z=4+3i-4λi+3λ=(4+3λ)+(3-4λ)i. Then we take the modulus: |z|=|(4+3λ)+(3-4λ)i|=sqrt((4+3λ)^2 + (3-4λ)^2 )=sqrt(16+24λ+9λ^2 + 9-24λ+16λ^2 )=sqrt(25+25λ^2 ). We want |z|=45, so we solve 45=sqrt(25+25λ^2 )=5*sqrt(1+λ^2 ), which gives us 1+λ^2 =9^2 , i.e. λ=±sqrt(80)=±4√5. That was kind of long and ugly though. Let's do a nicer method.

2. **"modulus, then multiply"**: Okay, this is a much faster way that uses the fact that |zw|=|z||w|. We know |z|=|(1-λi)(4+3i)|=|1-λi||4+3i|, and |4+3i|=sqrt(4^2 + 3^2 ) = 5, so we need to solve 45=5|1-λi|, which is |1-λi|=9. Much better! This immediately gives 1+λ^2 = 9^2 , which is exactly what we had above.

**Moral of the story:** multiplying real numbers is way easier than complex numbers, so using |zw|=|z||w| converts complex multiplication to real multiplication, and means we only have to multiply real numbers.

A lot of maths is like this: taking something scary or laborious to calculate, and using nice properties/theorems to get it to something simpler that we can do easily.
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(4 + 3i) has a magnitude of 5, so (1 - lambda i) must have a magnitude of 9 (since  the magnitude of a complex number is the product of its factors)
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What exactly is not clear to you?
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1)Multiply out the brackets

2)Sort out your expression for z in the form z=a+bi

3)Square root(a^2 +b^2 ) is equal to 45
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Let me guess, A level further maths? I did a core A level today
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Just multiply and find the modulus and equate them , what's not there to understand ?
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What class is this? Sounds terrible

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