# Cambridge Complex Numbers Question (1 Viewer)

#### jjjcjjj892

##### New Member
Just picked up 4 unit and we have started with complex numbers, so far its fairly easy but still trying to understand the new/abstract concepts.

Any help with the following question is greatly appreciated

z has modulus r and argument (theta). Find in terms of r and (theta) the modulus and one argument of:

c) iz

#### jazz519

##### Moderator
Moderator
modulus is given by sqrt(a^2 + b^2) where those values come from a + ib.

So in the above b=z, a=0

So modulus is z

For the argument(theta) draw out the complex number on an axis of a and b(this one being imaginary)

Find the angle formed through trig rules

In this case it’s just Pi/2

#### jjjcjjj892

##### New Member
Cheers for the response So just letting z1 = 0 + iz, you consider z to be the imaginary part of the expression and not as i(a+ib)? If yes then it makes sense that as there is no real part of z1 the angle must be pi/2 is this correct?

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#### jathu123

##### Active Member
$\bg_white \noindent Modulus of iz = |iz| = |i| \cdot |z| = r \\ (as the modulus of i is 1 and z is r) \\ Argument of iz = \arg (iz) = \arg (i)+ \arg(z) = \theta + \frac{\pi}{2} \\ (as the argument of i is \frac{\pi}{2} and z is theta)$

if you havent learnt the above stuff, then let z = a+ib, so iz = i(a+ib) = -b+ia. Now using the modulus formula, |iz| = sqrt( (-b)^2 +(a)^2 ) = r
Note that multiplying a complex number by i means thats you are rotating it by pi/2 radians anticlockwise, hence the new argument will be the argument of z + pi/2, which is theta + pi/2

Edit: z1 = a+ib, where 'a' and 'b' are purely real numbers, so when z1=0+iz, b cannot be z (as z is obviously a non real number). So you need to expand it instead.

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#### raz999

##### New Member
yep, jathu's spot on.
Jazz...not so much

#### 1729

##### Active Member
Just picked up 4 unit and we have started with complex numbers, so far its fairly easy but still trying to understand the new/abstract concepts.

Any help with the following question is greatly appreciated

z has modulus r and argument (theta). Find in terms of r and (theta) the modulus and one argument of:

c) iz
Multiplying by i is just a geometric anticlockwise rotation of pi/2, there is no enlargement or dilation so iz will also have modulus r.

As above, an argument would be pi/2 + theta

#### jjjcjjj892

##### New Member
Ah makes sense, cheers for the response, I had thought of splitting the mod of iz up like that but was confused with what to do with mod(i), only now do I realise its 1, seems so simple now ahah thanks for making me see that