Complex Number on Cartesian Coordinate System Question

I found this problem in a SAT Math II book and was confused by it:

In this figure, if the point Z represents a complex number a+bi, which of the points could represent i · z?

The figure has a x and y axis, which was confusing to me because I was under the impression that graphing complex numbers on a Cartesian coordinate system requires one to have x and a yi (imaginary axis). Further, I am not sure how one would multiply a complex number with a regular number and yield a point on the graph.

The answer is $D$ because multiplying by $i$ is equivalent to rotating around the origin with an angle of $\pi/2$.

• I'm unaware of that property – is there a proof that explains why that is so? – Shrey Apr 9 '15 at 21:30
• @ShreyDesai $(a + bi)* i = -b + a i$, which in the language of 2D vectors is the same as multiplying by the matrix $\left ( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right )$, which is the rotation matrix for a $\pi/2$ rotation about the origin. – John Colanduoni Apr 9 '15 at 22:06

$(a + bi) \cdot i = -b + ai$

So the resulting point $z \cdot i$ has:

1) An imaginary part which is the same as the real part of $z$ (slightly negative).

2) A real part which is the same as the negative of the imaginary part of $z$ (very negative).

Point D has these properties.

If you look at where $Z$ has been placed on the graph then you can conclude that since $Z=a+bi$ then $a$ must be negative and $b$ must be positive.

Therefore $iZ=i(a+bi)=ai+bi^2=ai-b=-b+ai$.

So we can conclude that the new point must lie to the left of the origin on the real axes (since we know that $b$ was positive).

similarly we can conclude that the new point must lie below the origin on the imaginary axes (since we know that $a$ is negative).

Now just look at the relative magnitudes of $a$ and $b$ from the original point $Z$ and you should be able to deduce the answer.